Frequently Asked Questions
Alex Lopez-Ortiz
This is a list of frequently asked questions for sci.math (version 1.4).
Any contributions/suggestions/corrections are most welcome.
(Thanks to all the people who already contributed).
Please use * e-mail * on any comment concerning the FAQ list.
Does anybody has references to these (or any of the following) results?
If so, please e-mail them to me (alop...@maytag.UWaterloo.ca)
so they can be added to the answers.
Q: What is the current status of Fermat's last theorem?
(There are no positive integers x,y,z, and n > 2 such
that x^n + y^n = z^n)
I heard that <insert name here> claimed to have proved it
but later on the proof was found to be wrong. ...
(wlog we assume x,y,z to be relatively prime)
A: The status of FLT has remained remarkably constant. Every
few years, someone claims to have a proof ... but oh, wait,
not quite. Meanwhile, it is proved true for ever greater
values of the exponent (but not all of them), and ties are
shown between it and other conjectures (if only we could
prove one of them), and ... so it has been for quite some time.
It has been proved that for each exponent, there are at most
a finite number of counter-examples to FLT, and (if you can
verify it) that there are counter-examples for at most
a finite number of exponents.
Q: Has the 4 colour theorem been solved?
(Every planar map with regions of simple borders can be coloured
with 4 colours in such a way that no two regions sharing
a non-zero length border have the same colour.)
A: This theorem was proved with the aid of a computer in 1976.
The proof shows that if aprox. 10,000 (?) basic forms of maps
can be colored with four colours, then any given map can be
colored with four colours So far nobody has been able to prove
it without using a computer. In principle it is possible
to emulate the computer proof by hand computations.
References:
K. Appel and W. Haken, Every planar map is four colourable,
Bulletin of the American Mathematical Society, vol. 82, 1976 pp.711-712.
K. Appel and W. Haken, Every planar map is four colourable,
Illinois Journal of Mathematics, vol. 21, 1977, pp. 429-567.
T. Saaty, The Four Colour Theorem: Assault and Conquest, McGraw-Hill, 1977.
Q: What are the values of:
largest known prime?
A: The largest known prime (currently) is 391581*2^216193 - 1.
See Brown, Noll, Parady, Smith, Smith, and Zarantonello,
Letter to the editor, American Mathematical Monthly, vol. 97,
1990, p. 214.
largest twin primes?
A: The largest known twin primes are 1706595*2^11235 +- 1.
See B. K. Parady and J. F. Smith and S. E. Zarantonello,
Largest known twin primes, Mathematics of Computation,
vol.55, 1990, pp. 381-382.
(Please send updates to alop...@maytag.UWaterloo.ca)
Q: I think I proved <insert big conjecture>. OR
I think I have a bright new idea.
What should I do?
A: Are you an expert in the area? If not, please ask first local
gurus for pointers to related work. If after reading them you still
think your *proof is correct*/*idea is new* then send it to the net.
Q: I have this complicated symbolic problem (most likely
a symbolic integral or an DE system) that I can't solve.
What should I do?
A: Find a friend with access to a computer algebra system
like MAPLE, MACSYMA or MATHEMATICA and ask her/him to solve it.
If none of this packages can solve it, then (and only then) ask
the net.
Q: Where can I get <Symbolic Computation Package>?
A: Maple Purpose: Symbolic and numeric computation, mathematical
programming, and mathematical visualization.
Contact: Waterloo Maple Software, 160 Columbia Street West,
Waterloo, Ontario, Canada N2L 3L3
Phone: (519) 747-2373 wm...@daisy.uwaterloo.ca wm...@daisy.waterloo.edu
A: DOE-Macsyma Purpose: Symbolic and mathematical manipulations.
Contact: National Energy Software Center
Argonne National Laboratory 9700 South Cass Avenue
Argonne, Illinois 60439 Phone: (708) 972-7250
A: Pari Purpose: Number-theoretic computations and simple numerical
analysis. Available for Sun 3, Sun 4, generic 32-bit Unix, and
Macintosh II. This is a free package, available by ftp from
math.ucla.edu (128.97.64.16). Contact: questions about pari
can be sent to pa...@mizar.greco-prog.fr
A: Mathematica Purpose: Mathematical computation and visualization,
symbolic programming. Contact: Wolfram Research, Inc.
100 Trade Center Drive Champaign, IL 61820-7237 Phone: 1-800-441-MATH
Q: What is 0^0 ?
A: According to some references, 0^0 is an undefined form
other books give it as 0 and a few others say it is 1,
the problem is that there are some good reasons for
choosing any of the three values.
Q: Why is 0.9999... = 1?
A: The Archimedean principle for the reals has as one of its
consequences (?) the fact 0.9999... = 1 .
Another argument could be given by noticing that the following sequence
of "natural" operations has as a consequence 1 = 0.9999....
Therefore it's "natural" to assume 1 = 0.9999.....
x = 0.99999....
10x = 9.99999....
10x - x = 9 9x = 9
x = 1
Thus
1 = 0.99999....
References:
E. Hewitt & K. Stromberg, Real and Abstract Analysis, Springer-Verlag,
Berlin, 1965.
W. Rudin, Principles of Mathematical Analysis, McGraw-Hill, 1976.
Q: Where I can get pi upto a few hundred thousand digits of pi?
Does anyone have an algorithm to compute pi to those zillion
decimal places?
A: MAPLE can give you 10,000 digits of Pi in a blink, and it can
compute another 20,000 overnight.
References from : jk8...@tut.fi
1. David H. Bailey
The computation of pi to 29,360,000 decimal digits using Borwein'
quartically convergent algorithm
Mathematics of Computation, Vol. 50, No. 181, Jan 1988, pp. 283-296
2. David H. Bailey
Numerical results on the transcendence of constants involving pi,
e, and Euler's constant
Mathematics of Computation, Vol. 50, No. 181, Jan 1988, pp. 275-281
3. P. Beckman
A history of pi
Golem Press, CO, 1971 (fourth edition 1977)
4. J.M. Borwein and P.B. Borwein
The arithmetic-geometric mean and fast computation of elementary
functions
SIAM Review, Vol. 26, 1984, pp. 351-366
5. J.M. Borwein and P.B. Borwein
More quadratically converging algorithms for pi
Mathematics of Computation, Vol. 46, 1986, pp. 247-253
6. J.M. Borwein and P.B. Borwein
Pi and the AGM - a study in analytic number theory and
computational complexity
Wiley, New York, 1987
7. Shlomo Breuer and Gideon Zwas
Mathematical-educational aspects of the computation of pi
Int. J. Math. Educ. Sci. Technol., Vol. 15, No. 2, 1984,
pp. 231-244
8. Y. Kanada and Y. Tamura
Calculation of pi to 10,013,395 decimal places based on the
Gauss-Legendre algorithm and Gauss arctangent relation
Computer Centre, University of Tokyo, 1983
9. Morris Newman and Daniel Shanks
On a sequence arising in series for pi
Mathematics of computation, Vol. 42, No. 165, Jan 1984,
pp. 199-217
10. E. Salamin
Computation of pi using arithmetic-geometric mean
Mathematics of Computation, Vol. 30, 1976, pp. 565-570
11. D. Shanks and J.W. Wrench, Jr.
Calculation of pi to 100,000 decimals
Mathematics of Computation, Vol. 16, 1962, pp. 76-99
12. Daniel Shanks
Dihedral quartic approximations and series for pi
J. Number Theory, Vol. 14, 1982, pp.397-423
13. David Singmaster
The legal values of pi
The Mathematical Intelligencer, Vol. 7, No. 2, 1985
14. Stan Wagon
Is pi normal?
The Mathematical Intelligencer, Vol. 7, No. 3, 1985
15. J.W. Wrench, Jr.
The evolution of extended decimal approximations to pi
The Mathematics Teacher, Vol. 53, 1960, pp. 644-650
--------------------------------------------------------------------------
Questions and Answers _Compiled_ by:
Alex Lopez-Ortiz alop...@maytag.UWaterloo.ca
University of Waterloo Canada
Dave Seaman
>Q: What is 0^0 ?
>
>A: According to some references, 0^0 is an undefined form
> other books give it as 0 and a few others say it is 1,
> the problem is that there are some good reasons for
> choosing any of the three values.
If we are going to have a FAQ list, it should at least give correct
information. There are at least two good reasons for 0^0=1, both following
from definitions of exponentiation in the standard literature.
The keeper of the FAQ list persists in claiming that there are good reasons for
choosing other values, though when I pressed him to give one, he was unable to
do so. Neither has anyone else on the net, though I have asked for reasons
before when the topic has come up.
Before everyone rushes to reply at once, let me point out that any argument
that depends on limits is fundamentally flawed. The function f(x,y) = x^y
has an essential discontinuity at the origin, which means that limits cannot be
used to find the value of 0^0. Moreover, such methods are guilty of circular
reasoning. Obviously, you cannot begin the defining process by using limits to
compute 1^1 or 3^2. You must begin with some more fundamental method. It
turns out that at least two fundamental methods exist, and both of them suffice
to define 0^0 along with all the other cases a^b where a and b are natural
numbers. Only after the function is extended to the rationals do limits come
into play. What justification is there to go back and change the
previously-defined 0^0 = 1 after exponentiation has been extended to the reals?
Some have suggested that 0^0 should be left undefined "to preserve continuity",
ignoring the facts that (1) discontinuous functions exist, and (2) even with
(0,0) removed from the domain, x^y remains discontinuous on the entire left
half plane (i.e. for x < 0), even though it is defined on a dense subset of
that half plane. Shall we decide that (-8)^(1/3) is undefined, instead of
being equal to -2, simply because the function is discontinuous there?
Unless someone is able to give a mathematically sound justification on the net,
I think the FAQ list should be amended to say that 0^0 = 1 and no justification
for any other value has been found.
--
Dave Seaman
a...@seaman.cc.purdue.edu
Kenneth Arromdee
>Unless someone is able to give a mathematically sound justification on the net,
>I think the FAQ list should be amended to say that 0^0 = 1 and no justification
>for any other value has been found.
I get the impression that the main "intuitive" reasons for 0^0 = 0 are the
following:
1) Preserving the identity 0^X = 0
2) One common "explanation" for why X^0 = 1 is that X^1 = X and then when you
divide by X, you get X^0 = 1, which would be division by zero for X=0.
(No, I don't agree with either one, just stating an observation...)
--
"This theory of yours -- that painful memories can be surgically removed..."
"I can't share details... one of my colleagues might steal my idea."
--Brenda Starr, 12/25/90
Kenneth Arromdee (UUCP: ....!jhunix!arromdee; BITNET: arromdee@jhuvm;
INTERNET: arro...@cs.jhu.edu)
Dave Seaman
>I get the impression that the main "intuitive" reasons for 0^0 = 0 are the
>following:
>1) Preserving the identity 0^X = 0
It is not an identity. 0^X is undefined for X < 0, is zero for X > 0, and
is one for X = 0.
>2) One common "explanation" for why X^0 = 1 is that X^1 = X and then when you
>divide by X, you get X^0 = 1, which would be division by zero for X=0.
Division by zero never proves anything other than the fact that your methods
are incorrect. The correct definition of 0^0 does not even involve division of
any kind, let alone division by zero.
>(No, I don't agree with either one, just stating an observation...)
Fair enough, but I just want to make it clear that there still are no valid
reasons for saying 0^0 != 1.
--
Dave Seaman
a...@seaman.cc.purdue.edu
M. Edward Borasky
0^X is 0 for X REAL and X > 0
0^X is NOT DEFINED for X REAL and less than or equal to 0
X^0 is 1 for X REAL and X not equal to 0. X^0 is NOT DEFINED for X REAL and equal to 0.
Therefore, 0^0 is NOT DEFINED. If it is not defined, then it is NOT
equal to 1. So
0^0 != 1
is true.
0^0 != 0
is true.
Any questions?
Dave Seaman
>Last time:
Good. That's one less person who will be arguing in the future.
>0^X is 0 for X REAL and X > 0
Correct, so far.
>0^X is NOT DEFINED for X REAL and less than or equal to 0
Wrong. 0^X is undefined for X REAL and < 0. The value of 0^0, by definition,
is the cardinality of the class of mappings from the empty set to itself, which
is perfectly well defined, and is precisely equal to 1.
I notice you didn't give a justification for your claim. I gave a
justification for mine. I also have a second justification, which I will hold
in reserve for now. You have some catching up to do.
>X^0 is 1 for X REAL and X not equal to 0. X^0 is NOT DEFINED for X REAL and equal to 0.
Again, you failed to give any justification for your claim. I still have my
two justifications for 0^0 = 1, and they are from the standard mathematical
literature.
>Therefore, 0^0 is NOT DEFINED. If it is not defined, then it is NOT
>equal to 1. So
Talk about begging the question. In case you didn't notice, I asked for a
mathematically sound reason to conclude that 0^0 != 1. You have not even
attempted to give a reason of any kind, sound or unsound. At least the
divide-by-zero crowd and the take-the-limit crowd is one step ahead of you.
Their arguments may be flawed, but at least they have arguments.
>0^0 != 1
>
>is true.
>
>0^0 != 0
>
>is true.
>
>Any questions?
Yes. Where did you learn logic?
I am still waiting to see a reason for 0^0 != 1.
--
Dave Seaman
a...@seaman.cc.purdue.edu
Herb Brown
>reasons for saying 0^0 != 1.
>
>--
>Dave Seaman
>a...@seaman.cc.purdue.edu
Yes, indeedy do! I agree with you 103%.
One should not say that 0^0 is undefined. One should say that 0^0 is
INDETERMINATE
It may turn out to be undefined, or it may turn out to be -12. In other
words, it is indeterminate; additional work needs to be done to determine
what its value is.
Herb
--
----------------------------------------------------------------------------
Herb Brown Math Dept The Univ at Albany Albany, NY 12222 (518) 442-4640
hib...@leah.albany.edu or hib...@cs.albany.edu or hb...@ALBNYVMS.BITNET
----------------------------------------------------------------------------
Alex Lopez-Ortiz
References:
largest known prime?
What should I do?
Q: What is 0^0 ?
A: According to some references, 0^0 is an undefined form
other books give it as 0 and a few others say it is 1,
the problem is that there are some good reasons for
choosing any of the three values.
Q: Why is 0.9999... = 1?
Dave Seaman
>One should not say that 0^0 is undefined. One should say that 0^0 is
>
> INDETERMINATE
>
>It may turn out to be undefined, or it may turn out to be -12. In other
>words, it is indeterminate; additional work needs to be done to determine
>what its value is.
May I suggest some appropriate additional work? A good starting point is
almost any textbook on axiomatic set theory and the foundations of the real
number system.
Recall that a natural number in ZF is actually a set. In particular, the
number zero is the empty set. The successor of a natural number n is the set
n union {n}, so that 0 = {}, 1 = {0} = {{}}, 2 = {0,1} = {{},{{}}}, etc.
There are definitions for arithmetic on the natural numbers, based on set
theory (here X represents the Cartesian product of sets and # represents the
cardinality of a set):
addition: m+n = #((m X {0}) union (n X {1}))
multiplication: mn = #(m X n)
exponentiation: m^n = #{ f : f : n -> m }
In other words, m^n is given by the number of mappings from n to m. Exercise:
let m = 2 = {0,1} and n = 3 = {0,1,2}, and count the mappings having domain
{0,1,2} and range {0,1} to determine the value of 2^3.
Now, try it with m = n = 0. The value of 0^0 is the cardinality of the class
of mappings from the empty set to itself, which is one. The one mapping
belonging to the class has an empty graph.
Once arithmetic is defined on the natural numbers, it can be extended in a
natural way to the integers, then to the rationals, then the reals and the
complex numbers. The only stage that involves anything like limits is the
extension from the rationals to the reals (but not exactly--the process is
called "completion" and it only superficially resembles limits). One of the
givens in this process is that m^n for m, n, rational is already defined and is
not to be changed by the completion process, which merely extends the
definition to cases that were not previously covered. Therefore, the value of
0^0 is not changed. It remains 1, as always.
------------------------------------------------------------------
There is an entirely different way to define arithmetic that is based on
algebraic concepts, but it leads to the same conclusion 0^0 = 1. In general,
if M is a monoid with identity e, and if m is an element of M and n is a
natural number, then m^n makes sense and is equal to the product from i = 1 to
n of m. In particular, if n = 0, then the product is empty, yielding a result
of e, the identity element (see Serge Lang, _Algebra_, page 6, for a
justification of this). Taking M = the ring of integers, we get 0^0 = 1.
Having empty products equal to 1 is analogous to having empty sums equal to 0,
a fact which old-fashioned adding machines used to depend on.
--
Dave Seaman
a...@seaman.cc.purdue.edu
Dave Seaman
>Q: What is 0^0 ?
>
>A: According to some references, 0^0 is an undefined form
> other books give it as 0 and a few others say it is 1,
> the problem is that there are some good reasons for
> choosing any of the three values.
If there are good reasons, why is no one able to give one? I have posted two
good reasons for 0^0 = 1, and no one has yet posted a reason for any other
interpretation.
It is not surprising that many textbooks do not define 0^0. That is because
a careful definition of exponentiation overlaps with many different areas of
mathematics, at least some of which are likely to lie outside the scope of
whatever textbook you are looking at. Moreover, a great deal of mathematics
can be done without knowing what 0^0 is. The result is that most authors do
not find the definition of 0^0 to be sufficiently useful to warrant the time
and effort required to explain it. A good definition can be found in most
textbooks on axiomatic set theory and the foundations of the real number
system, as I have previously pointed out.
This discussion is becoming like the infamous Monty Hall probability puzzle.
People insist on believing that 0^0 is undefined or indeterminate, even in the
presence of direct evidence to the contrary. One by one, people understand the
argument and stop trying to defend the indefensible, but always there are new
people ready to take up the argument.
--
Dave Seaman
a...@seaman.cc.purdue.edu
Lambert Meertens
more frequently than the questions are asked.
Furthermore, (a) the answers given are in general not very satisfactory,
nor can they be since this would in most cases require a small treatise;
and (b) there are recurrent questions that are not on the list (and no, I
am not going to say what I find missing since the repeated message is
already far too long.
I therefore suggest that the whole business be dropped.
--
--Lambert Meertens, CWI, Amsterdam; lam...@cwi.nl
Dave Seaman
>I therefore suggest that the whole business be dropped.
I second the motion. It seems that the list is not having its desired effect.
Most questions on the list do not have simple answers, but giving answers that
are completely wrong, on a repeating basis, is unforgiveable.
--
Dave Seaman
a...@seaman.cc.purdue.edu
Jon Noring
joined this thread. Anyway, set,
f= x^x
Taking the logarithm of both sides yields,
log(f)= x*log(x)
Or,
f= x^x= exp[x*log(x)]
As x -> 0 (using a limit argument), the expression x*log(x) tends to zero,
therefore f tends to one. I vote that 0^0= 1. I'm sure that there are
good arguments for other definitions. Let me know if my analysis has any
serious mathematical gaffs - I'm just a lowly engineer trying to play with
the big boys..
Jon Noring
John Merrill
one.
Recall that undefined forms are forms arising as limits where the
value of the limit depends on the means by which the limit is
approached. (Eg. 0/0 or \infty - \infty.) 0^0 is one of these.
Consider the expression
\lim_{a \to 0} |a|^{{b} \over {ln |a|}}.
For any real constant, b, this limit is 0^0. A straightforward
computation using L'Hospital's rule shows that the limit exists and
equals e^b. Since e^b can assume any positive value (depending upon
what b is, of course), 0^0 must be construed to be an undefined form.
It is left as a fairly direct exercise to construct a form of 0^0
which evaluates to 0 or to \infty.
--
John Merrill / mer...@bucasb.bu.edu / harvard!bu.edu!bucasb!merrill
Dave Seaman
>I don't know if analyzing 0^0 using limits has been presented, since I've just
>joined this thread.
Limits are not relevant to the discussion. It would be circular reasoning.
The steps in definining exponentiation are:
1. Define exponentiation on the integers. There is more than one way
to do this, as I pointed out in a posting earlier today.
2. Extend the definition to the rationals.
3. Extend the definition to the reals.
4. Extend the definition to the complex numbers (but this is outside
the scope of the present discussion).
The result 0^0 = 1 is obtained in step 1, as I have pointed out before.
Limits are not even available as a tool until you get to step 3. Step 3 does
nothing to change the results already established in steps 1 and 2. It merely
extends the domain.
Dave
--
Dave Seaman
a...@seaman.cc.purdue.edu
Dave Seaman
>Dave Seaman asks for a proof that 0^0 is an undefined form. Here's
>one.
>
>Recall that undefined forms are forms arising as limits where the
>value of the limit depends on the means by which the limit is
>approached. (Eg. 0/0 or \infty - \infty.) 0^0 is one of these.
Incorrect. The expression 0^0 does not call for the evaluation of a limit at
all. An example of an indeterminate form is
lim f(x)^g(x),
x -> a
where we are given that lim f(x) = lim g(x) = 0.
x -> a x -> a
If you think back carefully to elementary calculus, I think you will recall
that the phrase "indeterminate form" was never used except in connection with
the evaluation of limits, which has nothing at all to do with the present
discussion.
--
Dave Seaman
a...@seaman.cc.purdue.edu
Lambert Meertens
)
) INDETERMINATE
)
) It may turn out to be undefined, or it may turn out to be -12. In other
) words, it is indeterminate; additional work needs to be done to determine
) what its value is.
How much work would that be? Would 20 person years suffice? :-)
When the author of a mathematical text uses a mathematical symbol, it means
just what the author chooses it to mean -- neither more or less. So if you
choose to define + such that 2+2 = 5, fine, so be it; choosing the meaning
is your prerogative. It is therefore fundamentally impossible to
"determine" what the value of 2+2 is, except by using some given definition
of +. In particular, it is essentially, fundamentally, basically,
entirely, quite impossible to determine the meaning (value) of something if
its meaning is not fixed by (follows from) a given definition. Such a
thing cannot "turn out" to have some meaning after all.
However, if some existing definition for some symbol, say @, does not fix
the meaning of 3@7, say, you can always choose to *extend* the definition
of @ so as to assign to 3@7 any value you want! And even if it does fix a
meaning, you can modify it to give 3@7 a different meaning, or to leave 3@7
undefined.
Although I claim and am willing to defend that definitions *can* be chosen
arbitrarily, I strongly feel that they *should not* be chosen arbitrarily.
Getting the definitions "right" is of paramount importance. But what does
"right" mean, here? It is a matter of mathematical good taste to choose
definitions such that they are maximally useful. (As the context can (in
part) determine what is actually useful, it can govern the precise
definition.) It is a matter of sociological good taste to choose
definitions such that they do not unnecessarily diverge from commonly
accepted conventions, thereby jarring ingrained mathematical "reflexes".
(Occasionally the common conventions are not maximally useful; in such
cases I tend to let the mathematical usefulness prevail. In fact, the
mathematical conventions evolve in that direction.) I for one would
probably feel that defining 2+2 = 5 is wanting in taste in both respects.
If a tentative definition of @ leaves 3@7 undefined, an author should have
some specific good reason (like mathematical convenience) to assign a
particular value to it. But if such a good reason exists, and there are no
compelling arguments against it, it is good taste to extend the definition
as indicated.
Now for 0^0. This does, of course, not have some a priori meaning any more
than 1^1 or 0^(1/2) do. Any meaning it has has to follow from the
definition of ^. It is therefore completely pointless to argue what the
meaning of 0^0 *is*. You can of course argue what meaning, if any, it has
-- or argue that its meaning is not defined -- in the context of an
*already given* definition. But if you do this when it is the very
definition of 0^0 that is in doubt, you will run in circles, scream and
shout. Contributions to the discussion are only meaningful then if they
discuss what potential meaning is the most *useful* (including the option
of leaving this undefined).
From my own mathematical practice I know well that the definition 0^0 = 1
is useful. Any other possible definition is particularly useless. Leaving
it undefined is not only useless as well, but also (for careful authors) a
source of awkwardness or (for less careful authors) of minor errors. Thus,
I concur with Dave Seaman here: the "right" definition is: x^0 = 1, for
*all* x.
Some mathematicians, in defining ^, nevertheless choose, *for no good
reason*, to leave the meaning of 0^0 undefined. To do so, they even have
to go out of their way. I happen to think that this displays a lack of
mathematical taste.
Vincent Broman
> I have posted two
> good reasons for 0^0 = 1, and no one has yet posted a reason for any other
> interpretation.
"0^0 is indeterminate" means only this: viewing exponent(x,y) as a
function of two real variables, a value for exponent(0,0) cannot be
defined such that exponent(x,y) would be continuous at x=0, y=0.
In fact, for any nonnegative limit C you desire, real sequences a_n and b_n
can be chosen such that:
\lim_{n \rightarrow \infty} a_n = 0,
\lim_{n \rightarrow \infty} b_n = 0, and
\lim_{n \rightarrow \infty} a_n^{b_n} = C.
In all the following examples a_n^{b_n} = C for all n.
Let C = 0, then a_n = 0 and b_n = 1 / n.
Let C = 1, then a_n = 1 / n and b_n = 0
or for any other positive C,
a_n = e^{-n | \ln C |} and b_n = - \sign( \ln( C )) / n.
Vincent Broman, code 632, Naval Ocean Systems Center, San Diego, CA 92152, USA
Phone: +1 619 553 1641 Internet: bro...@nosc.mil Uucp: sdcsvax!nosc!broman
Lambert Meertens
) one.
) Recall that undefined forms are forms arising as limits where the
) value of the limit depends on the means by which the limit is
) approached. (Eg. 0/0 or \infty - \infty.) 0^0 is one of these.
I don't recall that terminology (:- possibly because my textbooks were not
in English :-), but in any case it is misleading. For example, define
f(x) = sum_{n=1}^infty sin(n*x)/n.
Now lim_{x uparrow 0} f(x) = -pi/2, and lim_{x downarrow 0} f(x) = +pi/2,
which is different. So, if understand the poster right, f(0) is an
undefined form. However, by the definition of f,
f(0) = sum_{n=1}^infty sin(n*0)/n
= sum_{n=1}^infty 0,
which, I hope, everyone will concede is well defined. Thus, here we have
an undefined form which is defined.
Consider next the definition
g(x) = 0/x.
Here lim_{x->0} g(x) = 0. This does not depend on the way the limit is
approached. Thus g(0) is not an undefined form. But, clearly, g(0) is
undefined.
Since people might easily be led to think that being an undefined form has
some relationship with a form being undefined -- which, apparently, is not
the case -- I consider "undefined form" severely misleading terminology
that should be avoided (and particularly in textbooks).
Vincent Broman
definitions get chosen. That is appreciated. When he opined that
there was never a good reason for leaving 0^0 undefined, I think
he swung a little wildly, even though I agree that most of the
time 0^0 could usefully be defined as 1.
In the arithmetic of constructive reals, discussions about limits and
the continuity of functions are vital. The value of a function at a
point in its domain where it is discontinuous CANNOT be computed from
a converging sequence of approximations to the function argument.
Attempts to do so will produce error bounds that do not tend to zero.
For similar reasons, it is dangerous to evaluate a floating point
function at a point where the (R->R) function is discontinuous.
The arguments were probably computed elsewhere using techniques
that involved roundoff, underflow, etc, and must be viewed as
inexact. In implementing an exponent function of two real arguments
for some math library, the case 0.0^0.0 should probably be excluded
from the domain of the function for two reasons:
1. The implementor of the function cannot possibly know which of
the arguments may or may not have underflowed to 0.0, or suffered
some other kind of approximation error; only the caller of the function
(actually the programmer) would know whether the zeroes really were
exact zeroes. So, he should supply the function value himself.
2. In the grand majority of cases where (0.0,0.0) is passed to
an exponent function, the programmer is doing something dimwitted,
like evaluating polynomials using one call to exponent for each term
of the polynomial. He can get a clue when it dumps core.
Adam David
+From: a...@seaman.cc.purdue.edu (Dave Seaman)
+Message-ID: <30...@mentor.cc.purdue.edu>
+
+In article <1990Dec24....@maytag.waterloo.edu> alop...@maytag.waterloo.edu (Alex Lopez-Ortiz) writes:
+
+>Q: What is 0^0 ?
+>
+>A: According to some references, 0^0 is an undefined form
+> other books give it as 0 and a few others say it is 1,
+> the problem is that there are some good reasons for
+> choosing any of the three values.
^^^^^
Can an undefined form have a value? That cannot be true, though some undefined
forms may have solution sets (possibly restrictable to single values if the
context permits).
+If we are going to have a FAQ list, it should at least give correct
+information. There are at least two good reasons for 0^0=1, both following
+from definitions of exponentiation in the standard literature.
Any reason to let 0^0 != 1 is going to have to be *very* good.
+Before everyone rushes to reply at once, let me point out that any argument
+that depends on limits is fundamentally flawed. The function f(x,y) = x^y
+has an essential discontinuity at the origin, which means that limits cannot be
+used to find the value of 0^0.
This depends on whether x=x. The function f(x) = x^x is demonstrably continous
(differentiable) at x=0, and in fact continuous for all x is Real, x^x is
Complex. (I have yet to see how z^z, z is Complex behaves, in case anyone was
wondering). Any limits taken on f(x) = x^x will give actual function values
(which are complex conjugate pairs for x < 0, x is non-integer). Negative
integer x give x^x Real, negative if x odd, positive if x even.
+Unless someone is able to give a mathematically sound justification on the net,
+I think the FAQ list should be amended to say that 0^0 = 1 and no justification
+for any other value has been found.
Yes, I'll second that.
+From: a...@seaman.cc.purdue.edu (Dave Seaman)
+Message-ID: <30...@mentor.cc.purdue.edu>
+
+In article <10...@emanon.cs.jhu.edu> arro...@cs.jhu.edu (Kenneth Arromdee) writes:
+
+>I get the impression that the main "intuitive" reasons for 0^0 = 0 are the
+>following:
+>1) Preserving the identity 0^X = 0
+
+It is not an identity. 0^X is undefined for X < 0, is zero for X > 0, and
+is one for X = 0.
0^x == |(x <= 0)| / |(x >= 0)|
x^0 == 1 if x >= 0, ||x^0|| == 1 for all Real x
+From: a...@seaman.cc.purdue.edu (Dave Seaman)
+Message-ID: <30...@mentor.cc.purdue.edu>
+
+In article <15...@ogicse.ogi.edu> bor...@ogicse.ogi.edu (M. Edward Borasky) writes:
+
+
+>0^X is 0 for X REAL and X > 0
+
+Correct, so far.
+
+>0^X is NOT DEFINED for X REAL and less than or equal to 0
+
+Wrong. 0^X is undefined for X REAL and < 0. The value of 0^0, by definition,
+is the cardinality of the class of mappings from the empty set to itself, which
+is perfectly well defined, and is precisely equal to 1.
Very good point.
Also:
If x=0 then it makes no sense looking at 0^x or x^0. x cannot be both constant
and variable at the same time.
+I notice you didn't give a justification for your claim. I gave a
+justification for mine. I also have a second justification, which I will hold
+in reserve for now. You have some catching up to do.
+
+--
+Dave Seaman
+a...@seaman.cc.purdue.edu
Was that complex numbers or a third one? I also have to catch up.
And I definitely agree that the FAQ list should not proclaim blatant untruths.
That way the subject may never be laid to rest.
I make the bold suggestion that Dave Seaman's second proof (witheld at present)
is simple enough to be worthy of inclusion in the FAQ list, as is the first one
already stated. Let's see it Dave, I'm probably not the only one curious.
Adam David. ad...@rhi.hi.is
Adam David
>0^x == |(x <= 0)| / |(x >= 0)|
>x^0 == 1 if x >= 0, ||x^0|| == 1 for all Real x
These two lines were intended to be flagged with a "garbage warning". The second
line is probably only partially correct, and neither line is directly relevant
to the question: x=x, 0=0, what is x^x at x=0?
Adam David.
ad...@rhi.hi.is
foster@groucho
arrangements of n items from a group of k items with replacement. So 0^0
is the number of arrangements of zero items from zero (with replacement). There
is exactly one such arrangment, namely the empty one (do nothing). So, 0^0=1.
James A. Foster
"Nothing exists"
------
Timothy Yi-chung Chow
>Unless someone is able to give a mathematically sound justification on the net,
>I think the FAQ list should be amended to say that 0^0 = 1 and no justification
>for any other value has been found.
O.K., I'll bite.
First, I agree that in the vast majority of cases, 0^0 should be
defined to equal one. Moreover, I can think of no context in which 0^0
should be defined to equal anything else. However, I think that there
is one context, namely in beginning courses in analysis, where it is
justifiable to leave 0^0 undefined.
What is my justification for this? First, note that in such courses a
set-theoretical definition of exponentiation is often inappropriate, if
for no other reason that it is probably beyond the ken of the students
at this level. Hence some other definition needs to be adopted. For
example, in most courses in complex analysis, exp(z) is defined in
terms of a power series, and then a^b is defined to be exp(b*log(a)) if
a!=0. Alternative approaches are possible, but all the common ones
require 0^0 to be treated as a special case that needs to be defined
separately. Now, in such courses there is no use for the definition
0^0=1. So why make such a definition? IMHO, definitions should only
be made if they are going to be used. So in this context, 0^0 is best
left undefined.
In fact, I can't think of any situation in analysis where the
definition 0^0=1 is useful. Mr. Seaman?
--
Tim Chow tyc...@phoenix.princeton.edu
Nowadays the only unforgivable sins are racism, sexism, homophobia, and bigotry.
Ger Timmens
>I was always satisfied with a combinatorial explanation. k^n is the number of
>arrangements of n items from a group of k items with replacement. So 0^0
>is the number of arrangements of zero items from zero (with replacement). There
>is exactly one such arrangment, namely the empty one (do nothing). So, 0^0=1.
>
Lim x^0 = 1, ==> 0^0 = 1.
x->0
However,
Lim 0^y = 0, ==> 0^0 = 0.
y->0
So it just depends on how you're approaching zero.
Happy New Year,
+===========================================================+
| Ger Timmens, 3 Dimensional Computer Vision Research Group |
| University Hospital Utrecht, Heidelberglaan 100 |
| 3584 CX Utrecht, THE NETHERLANDS, e-mail: g...@cv.ruu.nl |
| Tel.No.: +31-30-506711, Fax.No.: +31-30-513399 |
+-----------------------------------------------------------+
|> Anything you can do, I can do better. No you can't... <|
+-----------------------------------------------------------+
| Delft, University of Technology |
| Faculty of Technical Mathematics and Informatics |
| Section of Applied Analysis |
+===========================================================+
Herb Brown
>
>In fact, I can't think of any situation in analysis where the
>definition 0^0=1 is useful. Mr. Seaman?
>--
>Tim Chow tyc...@phoenix.princeton.edu
>Nowadays the only unforgivable sins are racism, sexism, homophobia, and bigotry.
Ah yes, you are certainly correct. Let's look at an example.
Consider the function f(x) = x^(1/ln(x)) for x in [0,1].
What does its output look like? Well, if we restrict x to (0,1],
then the graph looks 'good', i.e. it's connected. Can we set its
value at 0 so that it remains 'good' (connected, continuous) on
all of [0,1]? Well, let's see. Hmmm, if we substitute 0 for x we
obtain f(0) = 0^0, which, according to David Seamon & Michael Somos
should be 1. BUT, for values of x 'near' 0, we see that the output
f(x) is 'near' e (base of natural logs). So, to make our output
look 'good' it is reasonable to set f(0) = e (rather than 1).
As further justification, simply compute the limit of f(x) as x
approaches 0 from the right and you obtain e (not 1). Thus, in
our little example, 0^0 = e permits us to construct a 'nice looking'
graph on the entire interval [0,1], whereas assigning f(0) = 1
results in a broken (non-connected) output.
Mike Marcel Jonkmans
@In fact, I can't think of any situation in analysis where the
@definition 0^0=1 is useful. Mr. Seaman?
@--
@Tim Chow tyc...@phoenix.princeton.edu
@Nowadays the only unforgivable sins are racism, sexism, homophobia, and bigotry.
Consider series :
oo
exp (x) = Sum x^n / n! = x^0 + x^1 + x^2 / 2 + ....
n = 0
Now consider exp (0), it has the value 1.
The series gives us : 0^0 + 0^1 + 0^2 + .... = 0^0.
So in analysis 0^0 is defined as 1.
--
Mike Jonkmans. (mi...@cs.vu.nl)
..!uunet!cs.vu.nl!mike
Graeme Williams
>Consider series :
>
> oo
>exp (x) = Sum x^n / n! = x^0 + x^1 + x^2 / 2 + ....
> n = 0
>
>Now consider exp (0), it has the value 1.
>The series gives us : 0^0 + 0^1 + 0^2 + .... = 0^0.
>
>So in analysis 0^0 is defined as 1.
To my mind this is a somewhat spurious argument, the above is
a Taylor expansion namely
exp(x) = exp(0) + exp'(0) x + exp''(0) x^2/2 + exp'''(0) x^3/6 + ...
Placing x=0 in this requires only that exp(0)=1
The definition 0^0=1 is *only* needed to make the notation nice, that is,
to allow one to write all terms in the series as a summation. It is a
simple matter to remove the first term from the summation and by so
doing remove any NEED for defining 0^0=1.
IMHO whatever one decides 0^0 is, is purely a matter of convention.
One can for instance define any number of functions
f: f(x,y)=x^y except when x=y=0, in which case f(x,y)=k some constant.
The particular function you wish to call "exponentiation" is a matter
of *convention*.
Graeme Williams
gcwil...@watdragon.waterloo.edu
John McCarthy
is the most useful convention, because there is no convention
giving a particular value that makes x^y continuous at (0,0).
Indeed as x and y approach 0, x^y can be made to approach
any value you like.
Russell Turpin
"Most useful" for what??
What is or is not useful depends on what one is doing and what
one's purpose is. This whole discussion has bordered on the
ridiculous. Pace Mr Seagate, there have not been any mathematical
arguments presented in this thread. He has not produced a proof
that 0^0=1, but only stated his esthetic prejudices. Pace Mr
McCarthy, one cannot decide usefulness outside a particular context.
Mathematicians in this day and age should know better than to
argue over what definitions are "right". They should also
recognize the futility of arguing the esthetics or practicality
of a definition except in the context of a particular
presentation, subject area, or application.
Russell
Lambert Meertens
(Timothy Yi-chung Chow) writes:
) In fact, I can't think of any situation in analysis where the
) definition 0^0=1 is useful.
A simple example: Courant (who is a careful author) in his Differential and
Integral Calculus asserts several times (in Volume 1, second edition, on
pages 94 and 139) that
d/dx x^n = n*x^(n-1)
for every positive integer n. This is incorrect unless 0^0 is defined to
be 1.
Series expansions have been mentioned already. If you do actual (symbolic)
calculations with power series, it is far more convenient to be able to work
with expressions of the form
sum_{i in N} c_i*x^i
then having to wrestle with forms like
c_0 + sum_{i in N\{0}} c_i*x^i.
The greater convenience shows the usefulness of having 0^0 = 1.
Daniel E. Platt
Hi,
The problem with 0^0 = 1 is that it depends strongly on where you start, and
depends on what you're willing to bring to the problem (what context).
Algebraicly, x^0 is x/x. If x = 0, you're asking what number times 0 yields 0.
ALL numbers times 0 yield 0, so, x/x = 1 for x != 0, else its undefined.
In this context, strictly speaking, 0^0 is undefined.
Analysis is another way people approach this problem. Depending on how you
take limits you'll see the desired results. For example:
lim_{x->0} x^0 = 1.
Other limits could be
lim_{x->0+} x^x = 1.
which also works. An example of something that DOES NOT work is
lim_{x->0+} 0^x = 0.
In particular, in the plane:
lim_{x,y -> 0+} x^y = ?
depends on how the limit is evaluated.
Lastly, combinatorics would suggest for consistancy's sake that, to preserve
the consistancy of a particular formula, that 0^0 = 1... but that is just a
special case of a special formula (in my book at least...).
The original question was WHY are there so many disparate answers... the
answer to why there are so many is that there are so many ways that the
question can be answered, and many of them depend on particular contexts.
Hope this helps.
Dan
Dave Seaman
>First, I agree that in the vast majority of cases, 0^0 should be
>defined to equal one. Moreover, I can think of no context in which 0^0
>should be defined to equal anything else. However, I think that there
>is one context, namely in beginning courses in analysis, where it is
>justifiable to leave 0^0 undefined.
I agree that the author of a textbook has the right to decide what are the
necessary prerequisites for reading his book, and that the prerequisites
needed for a careful definition of exponentiation may well be outside the
scope of the text. I have no problem with authors evading the issue on
those grounds, though I wish they would either avoid mentioning 0^0
altogether or else use a neutral expression such as, "We do not assign a
value to 0^0." What they usually say, instead, is something like, "No
useful value can be assigned to 0^0." I suppose the statement is
technically correct: you don't ASSIGN a value to 0^0. You define
exponentiation by a general rule and then notice what the consequences are
when the rule is applied to the case 0^0. It may be possible to state a
general rule in such a way that it does not apply to the 0^0 case, but it
seems you have to go out of your way to do so, and the result is not as
simple as the rules that do apply to 0^0.
By the way, there is at least one other situation where 0^0 is undefined.
Let R be a ring without unity, and consider 0^0, where the exponent 0 is
the integer zero and the base 0 is the additive identity in R. By an
argument similar to that in Serge Lang's _Algebra_, it turns out that
r = 0^0 * r = r * 0^0
for every r in R, since 0^0 is just an empty product. But this would
imply that 0^0 is a unit element in R, contrary to hypothesis. Therefore,
I should probably modify my claim to say "0^0 = 1, provided 1 exists."
Even then, I can take the position that 0^0 = 1 because 0^0 is undefined
and 1 is undefined, and two undefined things are equal to each other.
(Don't laugh; I have seen similar statements in textbooks).
>In fact, I can't think of any situation in analysis where the
>definition 0^0=1 is useful. Mr. Seaman?
Lambert Merteens has already provided an excellent discussion of this.
I would add that you are placing the burden of proof on the wrong side.
I began this thread by asking for justifications for UNDOING the definition
of 0^0, since it needlessly complicates the issue and seems to serve no
purpose.
--
Dave Seaman
a...@seaman.cc.purdue.edu
Dave Seaman
>To my mind this is a somewhat spurious argument, the above is
>a Taylor expansion namely
>
>exp(x) = exp(0) + exp'(0) x + exp''(0) x^2/2 + exp'''(0) x^3/6 + ...
>
>Placing x=0 in this requires only that exp(0)=1
>The definition 0^0=1 is *only* needed to make the notation nice, that is,
>to allow one to write all terms in the series as a summation. It is a
>simple matter to remove the first term from the summation and by so
>doing remove any NEED for defining 0^0=1.
>
>IMHO whatever one decides 0^0 is, is purely a matter of convention.
>One can for instance define any number of functions
>
>f: f(x,y)=x^y except when x=y=0, in which case f(x,y)=k some constant.
>
>The particular function you wish to call "exponentiation" is a matter
>of *convention*.
You have it backwards. I am the one who wants to keep things simple. You
are the one who wants to introduce needless complications by making a
special case of 0^0. Therefore, I am the one who is entitled to ask for
justifications, which is exactly what I did when I started this thread.
My definition begins with:
If a and b are natural numbers, then a^b is the cardinality of the
class of mappings from b to a.
I don't know what your definition is, but apparently it begins with:
If a and b are natural numbers, then a^b is the cardinality of the
class of mappings from b to a, except that if a=b=0 we agree for no
good reason to ignore the natural definition and decree that a^b
has no value.
My version of the law of exponents for natural numbers is:
If a, b, and c are natural numbers, then a^(b+c) = a^b * a^c.
Your version of the law of exponents apparently must be:
If a, b, and c are natural numbers, and if either a is nonzero or
b*c is nonzero, then a^(b+c) = a^b * a^c.
The point of the latter circumlocution is that you have to avoid situations
like
0 = 0^1 = 0^(0+1) = 0^0 * 0^1 = undefined,
which shows that you created big trouble for yourself by singling out 0^0
and making it a special case.
The power series example shows yet again that your version of the definition
needlessly complicates things.
What do you have to gain by doing things in such a heavy-handed way? Did
you think you were rescuing x^y from the evils of discontinuity? Think
again. The function is discontinuous at infinitely many points of its
domain; for example, (-8) ^ (1/3) = -2, but in every neighborhood of
(-8,1/3) there are points (x,y) for which x^y > 0. All you have done, at
great cost in elegance and simplicity, is to remove ONE of the
discontinuities, a feat of truly monumental insignificance.
--
Dave Seaman
a...@seaman.cc.purdue.edu
Dave Seaman
>The problem with 0^0 = 1 is that it depends strongly on where you start, and
>depends on what you're willing to bring to the problem (what context).
True. I have stated before that there is more than one way to define
exponentiation on the integers, which can then be extended to the rationals and
the reals.
>Algebraicly, x^0 is x/x. If x = 0, you're asking what number times 0 yields 0.
>ALL numbers times 0 yield 0, so, x/x = 1 for x != 0, else its undefined.
>In this context, strictly speaking, 0^0 is undefined.
First of all, x^0 is x/x ONLY ON THE ASSUMPTION THAT x != 0. Therefore, you
can't use that equation to draw conclusions about 0^0. As I have pointed out
before, 0^0 is an empty product, which is equal to 1. Another example of an
empty product is 0! (zero factorial), which is 1.
>Analysis is another way people approach this problem. Depending on how you
>take limits you'll see the desired results. For example:
I think the problem is not "how you take limits" but how you apply the results.
>[...] An example of something that DOES NOT work is
>
> lim_{x->0+} 0^x = 0.
I don't know what you mean when you say that this example does not work.
Your computation is correct, and therefore it "works." Perhaps it's just that
you are trying to draw an unwarranted conclusion by assuming continuity at the
origin? The problem lies in your logic, not in the definition of
exponentiation.
>
>In particular, in the plane:
>
> lim_{x,y -> 0+} x^y = ?
>
>depends on how the limit is evaluated.
No. The limit you indicate does not exist. If you get a different result,
then your method of evaluating the limit is erroneous.
>Lastly, combinatorics would suggest for consistancy's sake that, to preserve
>the consistancy of a particular formula, that 0^0 = 1... but that is just a
>special case of a special formula (in my book at least...).
What do you mean by special case? Technically, 1^1, 3^2, and 5^7 are all
"special cases" of the definition for x^y. Why should 0^0 be any more or less
special than all the other special cases? And why is the formula any more
special than, say, the product rule for derivatives?
>The original question was WHY are there so many disparate answers... the
>answer to why there are so many is that there are so many ways that the
>question can be answered, and many of them depend on particular contexts.
It is true that the question can be answered in more than one way, but no one
has yet shown how to answer the question so that a result other than 0^0 = 1
is obtained. That is exactly the point of this discussion. When 0^0 = 1
arises in more than one way from possible definitions, why is there so much
resistance to adopting it? As I said at the beginning, "give me a reason."
--
Dave Seaman
a...@seaman.cc.purdue.edu
Gerald Edgar
(1) If a and b are cardinal numbers, then a^b is the cardinality of the
class of mappings from b to a.
According to this definition, 0^0 = 1. Reference: "Cardinal and Ordinal
Numbers", by W. Sierpinski, page 141 (second edition revised). Mathematical
credentials of the author: a leading Polish mathematician during the first
half of this century.
(2) Let a and b be ordinal numbers. Let Z(b,a) be the set of
all sequences (finite or transfinite) of type b whose terms
are ordinals numbers < a. Order Z(b,a) by ...[omitted]. Then
the order-type of Z(b,a) is the ordinal a^b.
Again, we get 0^0 = 1. Same Sierpinski reference, page 309. But actually,
he states the definition only for ordinals > 0. So technically, he leaves
0^0 undefined.
(3) If a and b are complex numbers, then a^b = exp(b log a), where
exp is the natural exponential and log is the natural logarithm.
This is the definition for complex numbers. According to this definition,
0^0 is undefined. Reference: "Complex Analysis" by L. Ahlfors, section 3.4.
Mathematical credentials of the author: Fields medalist.
So my own opinion is: either 0^0 = 1 or 0^0 is undefined, depending
on what kind of numbers you are talking about. Just writing "0" doesn't
tell you that.
--
Gerald A. Edgar
Department of Mathematics Bitnet: EDGAR@OHSTPY
The Ohio State University Internet: ed...@mps.ohio-state.edu
Columbus, OH 43210 ...!{att,pyramid}!osu-cis!shape.mps.ohio-state.edu!edgar
Dave Seaman
>(3) If a and b are complex numbers, then a^b = exp(b log a), where
> exp is the natural exponential and log is the natural logarithm.
>
>This is the definition for complex numbers. According to this definition,
>0^0 is undefined. Reference: "Complex Analysis" by L. Ahlfors, section 3.4.
>Mathematical credentials of the author: Fields medalist.
I am aware of the definition in Ahlfors, and I can see reasons for doing things
this way in the complex domain, which is the main reason that I said earlier in
this thread (when describing the steps involved in defining exponentiation)
that I considered complex numbers to be outside the scope of the present
discussion. However, I think there still are some problems to be addressed.
For example, you still want
d/dz z^n = n * z^(n-1)
in the complex domain, and this still remains untrue unless you have 0^0 = 1.
>So my own opinion is: either 0^0 = 1 or 0^0 is undefined, depending
>on what kind of numbers you are talking about. Just writing "0" doesn't
>tell you that.
It appears that even in the complex domain, you either want 0^0 = 1 or 0^0
undefined, depending on what you are trying to do.
--
Dave Seaman
a...@seaman.cc.purdue.edu
Lambert Meertens
ed...@shape.mps.ohio-state.edu (Gerald Edgar) writes:
) (3) If a and b are complex numbers, then a^b = exp(b log a), where
) exp is the natural exponential and log is the natural logarithm.
)
) This is the definition for complex numbers. According to this definition,
) 0^0 is undefined.
According to this definition, 0^b is undefined for *all* b.
Pramathanath Sastry
present discussion. What has been proved is :
\lim_{a \to 0} f(a)^{g(a)} is indeterminate.
In other words to find such a limit, a naive rule of the thumb is not
going to work, but what is needed is a less naive rule of the thumb
(for example, looking closer at f and g, and perhaps applying L'Hopital).
This is not the same as saying 0^0 is indeterminate. It is a nice symbol
and for reasons which must have been posted on this net (sorry I am a
very infrequent reader) I feel like assigning it a value equal to 1.
If it means that some functions
of the form f(x)^g(x), defined in a neighbourhood of zero, are not
continous at zero, so be it. I can live with that. And I can't
remember a single instance where I ever needed to know the value of
0^0.
Pramath Sastry.
migl...@cartan.math.nd.edu
> The point of the latter circumlocution is that you have to avoid
situations
> like
>
> 0 = 0^1 = 0^(0+1) = 0^0 * 0^1 = undefined,
>
> which shows that you created big trouble for yourself by singling out
0^0
> and making it a special case.
I'm a bit confused. By the same sort of reasoning, should we also define
1/0 to be 1?
First, note 0 = 0^1 = 0^2. Then
0 = 0^1 = 0^(2-1) = (0^2)*(0^(-1)) = (0^1)*(0^(-1))
Have we also created big trouble for ourselves in making 1/0 undefined?
Your answer, presumably, would be that you were speaking only about
natural numbers. Unfortunately, it doesn't seem to be universally agreed
that 0 is a natural number. For example, the first sentence of Ireland
and Rosen's book "Elements of Number theory" defines the natural numbers
as {1,2,3,...}.
If you take this definition of natural numbers, then your definition of
a^b as the cardinality of the class of mappings from a set of b elements
to a set of a elements, for a and b natural numbers, doesn't define 0^0
(or 0^a or a^0). You have to define these separately. And you make your
definition based on your needs. If you want something that'll stick when
you try to extend to the non-negative reals, then you make 0^0 undefined.
If you want something that only makes sense in certain (admittedly
natural) contexts, as you seem to want, then make it 1 and say that when
you extend you get a discontinuity at 0. This is fine. There's no reason
NOT to do that.
But by the same token suppose someone wants to define 0/0 = 1 because he
is exclusively interested in the function f(x) = x/x. He has a right to
do that, but he shouldn't expect his definition to work for every context.
So to go back to 0^0, if I'm only interested in the function 0^x for x a
positive real, and I want to extend this to the non-negative reals, it
would make sense for me to define 0^0 to be 0. The fact that you have a
context in which it makes more sense to consider 0^0 to be 1 doesn't
concern me in the restricted context of my situation. So it all boils
down to a definition, WHETHER IT BE OF THE NATURAL NUMBERS OR OF THE
EXPRESSION 0^0, or indeed, of the exponentiation function. All these
definitions will agree "almost everywhere," but it's exactly in the fuzzy
areas that one has to be careful about which definitions one is using.
In any case, you have to admit that the fact that you can find contexts in
which 0^0 should be something other than 1, but you can't find any such
contexts for, say, 2^3 not being 8, makes 0^0 special. You can say that
the most natural definition for it is 1, but you can't simply say that it
"is" 1 (the way you can say that 2^3 IS 8).
Risto Lankinen
Now, don't any of you see, what we've got here ???
lim(x^0) != lim(0^x) <-> P != NP for the least possible problem space.
x->0 x->0
Ain't that what this's all about? :->
Terveisin: Risto Lankinen
--
Risto Lankinen / product specialist ***************************************
Nokia Data Systems, Technology Dept * 2 2 *
THIS SPACE INTENTIONALLY LEFT BLANK * 2 -1 is PRIME! Now working on 2 +1 *
replies: ri...@yj.data.nokia.fi ***************************************
Dave Seaman
>First, note 0 = 0^1 = 0^2. Then
>
>0 = 0^1 = 0^(2-1) = (0^2)*(0^(-1)) = (0^1)*(0^(-1))
>
>Have we also created big trouble for ourselves in making 1/0 undefined?
The difference is that 1/0 has no natural meaning and none can be assigned in
any consistent way. On the other hand, 0^0 does have a natural meaning, which
does not lead to any inconsistencies, and omitting this meaning makes things
needlessly complex.
>Your answer, presumably, would be that you were speaking only about
>natural numbers. Unfortunately, it doesn't seem to be universally agreed
>that 0 is a natural number. For example, the first sentence of Ireland
>and Rosen's book "Elements of Number theory" defines the natural numbers
>as {1,2,3,...}.
I think most people agree that the natural numbers are those described by the
Peano axioms, which depend on the terms "zero," "number," and "successor."
>If you want something that'll stick when
>you try to extend to the non-negative reals, then you make 0^0 undefined.
In what way does 0^0 = 1 not "stick" when extended to the non-negative reals?
And why stop with the non-negative reals? Exponentiation is defined on an
uncountable, dense subset of the left half-plane, as well (and is discontinuous
at every such point, I might add).
In fact, I think this phobia over discontinuity has gone far enough. Everyone
is assuming that discontinuity is a Bad Thing, to be avoided at all costs
(including circular reasoning). Discontinuity is fairly small potatoes as
mathematical pathology goes, but judging by the controversy in this newsgroup,
it is not to be underestimated. I hereby take the position that pathology in
general is a Good Thing, indeed a thing to be sought after in mathematics.
Pathology is to mathematics as dissonance is to music; without it a vital
element would be missing, and things would be very boring.
Therefore, I now abandon all my previous arguments and declare that the
discontinuity of x^y at (0,0) is in itself sufficient justification to declare
that 0^0 = 1.
[Only partly :-), by the way]
>But by the same token suppose someone wants to define 0/0 = 1 because he
>is exclusively interested in the function f(x) = x/x. He has a right to
>do that, but he shouldn't expect his definition to work for every context.
But x/x becomes a very boring function if you fill in the hole at x=0. It is
only interesting when left as is. Why else would anyone bother to study it?
>In any case, you have to admit that the fact that you can find contexts in
>which 0^0 should be something other than 1, but you can't find any such
>contexts for, say, 2^3 not being 8, makes 0^0 special. You can say that
>the most natural definition for it is 1, but you can't simply say that it
>"is" 1 (the way you can say that 2^3 IS 8).
Again, you have it backwards. Let F be the field of integers modulo 5. Then
in F, we have 2^3 = 3 but still 0^0 = 1.
--
Dave Seaman
a...@seaman.cc.purdue.edu
Jonathan King
Doug Merritt
>
>Recall that undefined forms are forms arising as limits where the
>value of the limit depends on the means by which the limit is
>approached. (Eg. 0/0 or \infty - \infty.) 0^0 is one of these.
I thought that many of these things have well defined forms under
nonstandard analysis, which rather undermines your argument.
Doug
Doug Merritt do...@eris.berkeley.edu (ucbvax!eris!doug)
or uunet.uu.net!crossck!dougm
Doug Merritt
>
>This depends on whether x=x. The function f(x) = x^x is demonstrably continous
>(differentiable) at x=0, and in fact continuous for all x is Real, x^x is
>Complex.
Untrue; simply drawing the graph shows quite obviously that x^x tends towards
infinity as x approaches zero, and yet 0^0 != infinity. I.e. it is
not continuous at x=0. You're also apparently implying that "continuous"
is the same as "differentiable", which is not true. Many functions are
known which are everywhere continuous yet nowhere differentiable.
(I have yet to see how z^z, z is Complex behaves, in case anyone was
>wondering). Any limits taken on f(x) = x^x will give actual function values
>(which are complex conjugate pairs for x < 0, x is non-integer). Negative
>integer x give x^x Real, negative if x odd, positive if x even.
The function x^x for negative x is not differentiable (not sure whether
it's continuous); it has an infinite range of values for any given
transcendental value of x, and a finite number for rational x. In
particular it has up to q values for rational x = p/q.
I'm not sure how many values it has for x an algebraic irrational,
although if the answer were "infinite" then that would lead me to
think that there are aleph-sub-1 values for transcendental x.
I'm still foggy on how it behaves for complex x.
Doug Merritt
>
>Consider the function f(x) = x^(1/ln(x)) for x in [0,1].
>What does its output look like? Well, if we restrict x to (0,1],
>then the graph looks 'good', i.e. it's connected. Can we set its
>value at 0 so that it remains 'good' (connected, continuous) on
>all of [0,1]? Well, let's see. Hmmm, if we substitute 0 for x we
This is all bogus (sorry). Observe that, by the Complex Analysis
definition of exponentiation, x^y :== exp(y ln x).
Therefore, x^(1/ln(x)) = exp(1/ln(x) * ln(x)) by definition, which
simplifies to exp(ln(x) / ln(x)), which simplifies to exp(1),
which equals e, again by definition (of 'e' in Complex Analysis).
So you can't draw any conclusions from x^(1/ln x) at all; it is
exactly the function f(x) = e.
This is analogous to arguing about the functions f(x) = x(e/x),
which also equals e regardless of the value of x, once you simplify.
> So, to make our output
>look 'good' it is reasonable to set f(0) = e (rather than 1).
>As further justification, simply compute the limit of f(x) as x
>approaches 0 from the right and you obtain e (not 1). Thus, in
>our little example, 0^0 = e permits us to construct a 'nice looking'
>graph on the entire interval [0,1], whereas assigning f(0) = 1
>results in a broken (non-connected) output.
Hopefully it is now clear that this result demonstrates the nature
of the arbitrary function f(x) = e cast into exponential form, rather
than on the nature of exponential forms in general.
Doug Merritt
>uncountable, dense subset of the left half-plane, as well (and is discontinuous
>at every such point, I might add).
Why only a subset of the left half-plane? On which points is it not
defined?
Dave Seaman
>In article <31...@mentor.cc.purdue.edu> a...@seaman.cc.purdue.edu (Dave Seaman) writes:
>>
>>And why stop with the non-negative reals? Exponentiation is defined on an
>>uncountable, dense subset of the left half-plane, as well (and is discontinuous
>>at every such point, I might add).
>
>Why only a subset of the left half-plane? On which points is it not
>defined?
The value of x^y (where x and y are real) is undefined at points where x < 0
and y is either (a) irrational, or (b) rational, of form p/q in lowest terms,
where q is even.
Of course, you can assign complex values (many of them), but I am specifically
discussing the single-valued real function x^y here.
The value of x^y is defined when x < 0 and y = p/q in lowest terms, where q is
odd. Moreover, assuming x < 0 and q odd, we get x^(p/q) = (x^p)^(1/q) =
(x^(1/q))^p < 0 when p is odd, and x^(p/q) > 0 when p is even. This makes for
lots of discontinuities (uncountably many, in fact), which makes the single
discontinuity at (0,0) look very unimportant, indeed.
--
Dave Seaman
a...@seaman.cc.purdue.edu
Dave Seaman
>In article <1990Dec31....@zaphod.mps.ohio-state.edu>
>ed...@shape.mps.ohio-state.edu (Gerald Edgar) writes:
>)
>) (3) If a and b are complex numbers, then a^b = exp(b log a), where
>) exp is the natural exponential and log is the natural logarithm.
>)
>) This is the definition for complex numbers. According to this definition,
>) 0^0 is undefined.
>
>According to this definition, 0^b is undefined for *all* b.
An interesting sidelight on this is that if f(z) = z^0, which is defined
everywhere except at z=0 according to the above definition, then you can use
analytic continuation to show f(0) = 1.
You can't use the same reasoning for g(z) = 0^z, because the function is not
analytic. According to Lambert's observation, g is not defined anywhere, but
even if you are generous, I don't think g can be defined for anything other
than the nonnegative reals. In particular, suppose 0^i = w. Then w^i =
(0^i)^i = 0^(i*i) = 0^(-1) = undefined, implying log w is undefined.
Obviously, w cannot be zero, since then 0 = 0^i = w^i = (0^i)^i = 0^(i*i) =
0^(-1). Therefore, no finite value is possible for 0^i, and it follows that
0^(a+bi) = 0^a * (0^i)^b has no finite value when b != 0.
--
Dave Seaman
a...@seaman.cc.purdue.edu
Herb Brown
>>
>>Consider the function f(x) = x^(1/ln(x)) for x in [0,1].
>>What does its output look like? Well, if we restrict x to (0,1],
>>then the graph looks 'good', i.e. it's connected. Can we set its
>>value at 0 so that it remains 'good' (connected, continuous) on
>>all of [0,1]? Well, let's see. Hmmm, if we substitute 0 for x we
>
>
>simplifies to exp(ln(x) / ln(x)), which simplifies to exp(1),
>which equals e, again by definition (of 'e' in Complex Analysis).
>
>So you can't draw any conclusions from x^(1/ln x) at all; it is
>exactly the function f(x) = e.
>
>
Doug, you have not read this carefully. The function equals e only on the
interval (0,1]. It equals x^(1/ln(x)) on the entire interval [0,1]. The
problem occurs at the endpoint x = 0. At that point, f(0) becomes 0^0,
which according to the recent fuss being made should be set equal to 1.
I'm simply pointing out that if we follow this advice and set f(0) = 1
then our output is 'broken' at the left endpoint (which may be OK with
some people and not OK with others). For those that wish a 'smoother'
journey towards the left endpoint, they will need to set f(0) = e. Their
trip will now be 'nice' to them, meaning no disaster as they reach the
left endpoint. (This may be an important consideration in building roads,
for example.)
My illustration is simply an exercise from one of the seven indeterminate
forms.
Dave Seaman
>For those that wish a 'smoother'
>journey towards the left endpoint, they will need to set f(0) = e.
If e = 0^0, then e^2 = 0^0 * 0^0 = 0^(0+0) = 0^0 = e, which implies either e=0
or e=1. This probably will do a lot to simplify all those messy power series
expansions. You may be on to something here.
--
Dave Seaman
a...@seaman.cc.purdue.edu
migl...@cartan.math.nd.edu
(a) For two natural numbers b,c (including 0 since we're not singling out
0 these days) we say that b divides c if there exists a natural number a
such that ba = c. Hence 0 divides 0 since 0*1 = 0.
(b) Given two natural numbers b,c such that b divides c, we define c/b as
follows. Take a set consisting of c elements and partition it into
subsets each of cardinality b. Then c/b is the number of subsets required
for such a partition.
Therefore taking b = c = 0, we get 0/0 = 1. Or if you want you can
interpret this as giving 0/0 = 0. In any case, 0/0 should have a specific
value.
Or, alternatively, you can treat 0 as a special case in (a) and say that
0/0 is undefined. That's essentially what we're doing with 0^0.
Dave Seaman
>Maybe we should also decide that 0/0 = 1. Here's the reasoning.
You probably think I'm going to tell you that you can't do that. Well, you're
wrong. I'm going to explain how you CAN do that, and what the consequences
are.
>(a) For two natural numbers b,c (including 0 since we're not singling out
>0 these days) we say that b divides c if there exists a natural number a
>such that ba = c. Hence 0 divides 0 since 0*1 = 0.
This is quite reasonable, so far.
>(b) Given two natural numbers b,c such that b divides c, we define c/b as
>follows. Take a set consisting of c elements and partition it into
>subsets each of cardinality b. Then c/b is the number of subsets required
>for such a partition.
By definition, c/b means c * b^(-1), provided b has an inverse.
>Therefore taking b = c = 0, we get 0/0 = 1. Or if you want you can
>interpret this as giving 0/0 = 0. In any case, 0/0 should have a specific
>value.
Assume 0 has an inverse, 0^(-1). This means 0 * 0^(-1) = 1. On the other
hand, 0 * 0^(-1) must also equal 0, because 0 * x = 0 for all x. We conclude,
then, that 0 = 1.
Some might regard 0 = 1 as a contradiction, but actually it is not. It is a
true statement if you are dealing with the trivial ring having just one
element, and false otherwise.
Therefore, you can set 0/0 = 0, or 0/0 = 1. Both statements are equivalent,
and neither leads to a contradiction. However, a necessary consequence of your
decision is that there is only one number (zero, or one, or whatever you wish
to call it). You may certainly choose to do this, if you don't mind having a
very boring arithmetic.
>Or, alternatively, you can treat 0 as a special case in (a) and say that
>0/0 is undefined. That's essentially what we're doing with 0^0.
You do not need to treat 0 as a special case in (a) in order to conclude
that 0/0 is undefined. All that is necessary is to postulate that 0 != 1 (i.e.
more than one number exists), and it necessarily follows that 0/0 must be
undefined.
This is not at all equivalent to what we are doing with 0^0, which is fully
compatible with arithmetic on the real number system, and which arises in a
natural way from the basic definitions.
--
Dave Seaman
a...@seaman.cc.purdue.edu
migl...@cartan.math.nd.edu
mimic what you did, namely to start with the natural numbers (including 0
in our definition of natural numbers) and define natural operations in a
natural way, and see what consequences this gives. So I first define a
multiplication function *: NxN --> N (in the obvious way using the
disjoint union of c sets of cardinality b, or however you want to do it).
Now I want to define an operation / on NxN which, however, will only be
> >(a) For two natural numbers b,c (including 0 since we're not singling
out
> >0 these days) we say that b divides c if there exists a natural number
a
> >such that ba = c. Hence 0 divides 0 since 0*1 = 0.
>
> >(b) Given two natural numbers b,c such that b divides c, we define c/b
as
> >follows. Take a set consisting of c elements and partition it into
> >subsets each of cardinality b. Then c/b is the number of distinct
subsets required
> >for such a partition.
>
This is how I'm defining c/b from scratch (I've added the word "distinct",
which is only needed for the 0/0 case), using set-theoretic notions and
the natural numbers, in analogy with your definition of a^b as the number
of functions from a set of cardinality b to one of cardinality a. I don't
need you to tell me that 0 has no inverse in the rational numbers and in
fact only has an inverse in a field with one element. The point I'm
trying to make is that one can very well define a notion for the
non-negative natural numbers which needs to be treated gingerly when you
throw in 0 or when you try to extend it to the rationals or the reals.
Are you saying that (b) doesn't give 0/0 = 1? There's only one way to
write the empty set as a disjoint union of distinct sets of 0 elements,
namely A = A.
What you're telling me is that no, the definition of c/b isn't what I
gave. It is defined as the product of c and 1/b, where 1/b is the unique
rational number x with the property that x*b = 1, and that of all the
natural numbers, 0 is special and needs to be treated in a special way.
You're saying that my definition is fine as long as b != 0, and it agrees
with the "right" definition in this case, but even though the words make
sense when b = c = 0 you still have to treat this as a special case.
Which is precisely what I've been saying about 0^0. To paraphrase you in
a much earlier post, you're saying that the natural set-theoretic notion
of b/c as given by (a) and (b) is fine as long as c isn't 0, but it
doesn't hold for c = 0. So why is it so bad to say that no, a^b is
defined (for the non-zero natural numbers) as the product of a with itself
b times, and this agrees with your natural set-theoretic description. But
0 needs to be handled separately even though the words make sense.
Finally, to reply to your comment
>
> This is not at all equivalent to what we are doing with 0^0, which is
fully
> compatible with arithmetic on the real number system, and which arises
in a
> natural way from the basic definitions.
I have to say that to some of us, "fully compatible with arithmetic on the
real number system" includes the notion of continuity. (As a natural
consequence of passing from a discrete set N to a continuum R.) If a
positive sequence x(n) --> a and any sequence y(n) --> b then x(n)^y(n)
--> a^b as long as this latter expression is defined. Since I can find
x(n) and y(n) converging to 0 with x(n)^y(n) converging to any positive
value I want (which is a lot different from simply going off to infinity
or being undefined), I find it incompatible to give a^b any specific value.
You clearly don't find this incompatible, and I'm not saying you're wrong.
I'm just answering your original question about why someone might have a
problem with 0^0 = 1. Namely, one might want that if x(n),y(n) are
non-negative for each n and x(n) --> a and y(n) --> b and a^b is defined
then x(n)^y(n) --> a^b. That is, one might want compatibility not just
with arithmetic but with the fuller structure of the real numbers, to as
large degree as possible.
Dave Seaman
>Dave, I don't think you follow what I'm trying to say. I'm trying to
>mimic what you did, namely to start with the natural numbers (including 0
>in our definition of natural numbers) and define natural operations in a
>natural way, and see what consequences this gives.
I understood that you were trying to mimic what I did, and therefore I assumed
you were trying to construct a ring. I merely used the ring axioms to draw
some conclusions. I see now that you do not understand such things as rings
and fields, and therefore I will avoid them in this message. Herstein's
_Topics_in_Algebra_ would be a good place to start if you want to learn more.
>> >(b) Given two natural numbers b,c such that b divides c, we define c/b
>as
>> >follows. Take a set consisting of c elements and partition it into
>> >subsets each of cardinality b. Then c/b is the number of distinct
>subsets required
>> >for such a partition.
>>
>Are you saying that (b) doesn't give 0/0 = 1? There's only one way to
>write the empty set as a disjoint union of distinct sets of 0 elements,
>namely A = A.
Actually, there are two ways. I think this example quite aptly demonstrates
the difference in how we think. Even when dealing with empty sets, you
unconsciously assume that there must be at least one of everything.
Your collection consists of one set, which is empty. In symbols: {{}}.
My collection consists of no sets at all. In symbols: {}.
The union of your collection of sets (all one of them) is empty.
The union of my collection of sets (all zero of them) is also empty.
An empty union is analogous to an empty product, which is where this discussion
began in the first place. Exercise: in your favorite programming language,
write a loop to compute the product of a collection of numbers stored in an
array. Assume that the variable N gives the actual size of the collection.
I'll wait here while you do that.
Finished? Now, suppose N = 0. What does your code do? Perhaps you need to
rewrite it because you hadn't considered the possibility of computing an empty
product. You should wind up with some sort of loop statement that may be
executed zero times. Question: what value does your code compute for the
product when N is zero?
> ... you still have to treat this as a special case.
>
>So why is it so bad to say that no, a^b is
>defined (for the non-zero natural numbers) as the product of a with itself
>b times, and this agrees with your natural set-theoretic description. But
>0 needs to be handled separately even though the words make sense.
I gave two different definitions, and you seem to be combining elements of both
in the same sentence. However, I don't see what you mean about separate
handling and special cases. Special cases are what deductive logic is all
about. You reason from the general to the specific.
"Socrates is mortal" is a special case of the proposition "All humans
are mortal."
"Six is composite" is a special case of the proposition "all even
numbers greater than two are composite."
"0^0 = 1" is a special case of the proposition "x^0 = 1 for all x."
"3^0 = 1" is another special case of the same proposition.
I don't see anything here that requires special handling. All that is needed
is simple instantiation.
>I have to say that to some of us, "fully compatible with arithmetic on the
>real number system" includes the notion of continuity. (As a natural
>consequence of passing from a discrete set N to a continuum R.)
Oh, I see. Then you can't compute any limits of the form
f(x+h) - f(x)
lim -------------
h -> 0 h
since all you get is 0/0. Poof! There goes calculus.
Doesn't it seem redundant to you to have two different notations for
g(a) and lim g(x)
x -> a
if they are always supposed to mean the same thing? The study of limits is
precisely what real analysis is all about. If you learn some more mathematics,
you will find that continuity is a relative concept: it depends on what
topology you are using. The same function may be continuous with respect to
one topology, and discontinuous with respect to another.
>If a
>positive sequence x(n) --> a and any sequence y(n) --> b then x(n)^y(n)
>--> a^b as long as this latter expression is defined. Since I can find
>x(n) and y(n) converging to 0 with x(n)^y(n) converging to any positive
>value I want (which is a lot different from simply going off to infinity
>or being undefined), I find it incompatible to give a^b any specific value.
First, notice that (-8) ^ (1/3) = -2, because (-2)^3 = -8.
Notice also that (-8) ^ (2/3) = (-2)^2 = +4, not -4. In general, if x<0 and
y is a rational number p/q (expressed in lowest terms), then there are three
possibilities:
x ^ (p/q) does not exist if q is even, since there are no real even
roots of negative numbers.
x ^ (p/q) exists and is negative if p and q are odd, since the q-th
root of x is negative and an odd power of a negative is a negative.
x ^ (p/q) exists and is positive if p is even and q is odd, since an
even power of a negative is a positive.
Now, what is the limit of the following sequence?
(-8) ^ (10/31)
(-8) ^ (100/301)
(-8) ^ (1000/3001)
(-8) ^ (10000/30001)
.
.
.
Notice that every term in the sequence is positive. What does this tell you
about the continuity of x^y at the point (-8,1/3)?
As I have been pointing out in other postings, x^y has uncountably many
discontinuities in its domain. Why, then, is the one at (0,0) so important?
>You clearly don't find this incompatible, and I'm not saying you're wrong.
> I'm just answering your original question about why someone might have a
>problem with 0^0 = 1. Namely, one might want that if x(n),y(n) are
>non-negative for each n and x(n) --> a and y(n) --> b and a^b is defined
>then x(n)^y(n) --> a^b. That is, one might want compatibility not just
>with arithmetic but with the fuller structure of the real numbers, to as
>large degree as possible.
The only "problem" I can see with 0^0 = 1 is for people who are not comfortable
working with limits.
--
Dave Seaman
a...@seaman.cc.purdue.edu
migl...@cartan.math.nd.edu
don't think you intended to be insulting. I'll just make a few short
remarks.
>
> Your collection consists of one set, which is empty. In symbols: {{}}.
> My collection consists of no sets at all. In symbols: {}.
>
> The union of your collection of sets (all one of them) is empty.
> The union of my collection of sets (all zero of them) is also empty.
>
Fine. I'm happy to accept an argument for 0/0 = 0. I'm not arguing that
0/0 has any specific value-- on the contrary. I'm trying to say that you
need to handle 0 as a special case.
> >So why is it so bad to say that no, a^b is
> >defined (for the non-zero natural numbers) as the product of a with
itself
> >b times, and this agrees with your natural set-theoretic description.
But
> >0 needs to be handled separately even though the words make sense.
Incidently, this is the way Lang does it. For a multiplicative group G he
defines x^0 = e, x^n = x*...*x n times (for n positive) and x^(-n) =
(x^(-1))^n. You'll note that this doesn't have anything to do with 0^0
yet. I just want to point out that this is the standard definition.
>
> I gave two different definitions, and you seem to be combining elements
of both
> in the same sentence. However, I don't see what you mean about separate
> handling and special cases. Special cases are what deductive logic is
all
> about. You reason from the general to the specific.
>
> "Socrates is mortal" is a special case of the proposition "All
humans
> are mortal."
>
> "Six is composite" is a special case of the proposition "all
even
> numbers greater than two are composite."
>
> "0^0 = 1" is a special case of the proposition "x^0 = 1 for all
x."
> "3^0 = 1" is another special case of the same proposition.
>
> I don't see anything here that requires special handling. All that is
needed
> is simple instantiation.
Sure, just as 0/0 = 1 is a special case of "x/x = 1 for all real x". It's
your proposition which is in question, not your instantiation.
>
> As I have been pointing out in other postings, x^y has uncountably many
> discontinuities in its domain. Why, then, is the one at (0,0) so
important?
>
Thanks for asking the key question. It's so important because the
function x^y is continuous on the first quadrant minus the axes and it is
continuous if you extend this domain to the positive x and y axes. But
there's no way to extend it to the origin and preserve continuity. It's
important because it's in the closure of the natural domain of the
function, and especially because it doesn't just blow up there, it takes
on any positive value you like.
This is why, in any calculus book, you'll find the SYMBOL 0^0 representing
a particular type of indeterminate form. This is a symbol, not a number.
You're right-- it's just a convenient way of representing a particular
type of limit. Some people would say that this justifies saying that the
function isn't defined at the origin. If you want to say that the value
is actually 1 and you simply have a discontinuity, that's fine too.
Somewhere along the line you have to separate 0 from the rest of the
natural numbers, though. Because as you pointed out, if c is a natural
number (EXCEPT 0) and a,b are integers then c^(a+b) = (c^a)*(c^b). But
this rule would give 0^0 = 0^(1-1) = (0^1)*(0^(-1)) which is undefined.
Dave Seaman
>Incidently, this is the way Lang does it. For a multiplicative group G he
>defines x^0 = e, x^n = x*...*x n times (for n positive) and x^(-n) =
>(x^(-1))^n. You'll note that this doesn't have anything to do with 0^0
>yet. I just want to point out that this is the standard definition.
Then, a little later...
>Sure, just as 0/0 = 1 is a special case of "x/x = 1 for all real x". It's
>your proposition which is in question, not your instantiation.
Look again at what you just wrote above. You just agreed (quoting Lang) that
x^0=e for all x. So why are you saying that the proposition is in question?
You agreed that the general proposition is correct, and you agreed that the
instantiation is correct. What is your point?
>> As I have been pointing out in other postings, x^y has uncountably many
>> discontinuities in its domain. Why, then, is the one at (0,0) so
>important?
>Thanks for asking the key question. It's so important because the
>function x^y is continuous on the first quadrant minus the axes and it is
>continuous if you extend this domain to the positive x and y axes. But
>there's no way to extend it to the origin and preserve continuity. It's
>important because it's in the closure of the natural domain of the
>function, and especially because it doesn't just blow up there, it takes
>on any positive value you like.
Discontinuities are funny that way. They frequently occur on the
boundaries of regions of continuity, and they frequently are
associated with wild behavior near the singularity.
I have been reading your explanation over and over, trying very
hard to understand. Your entire argument sounds to me like, "I
don't like chameleons, because they are always changing colors."
I will resist the urge to ask, "Why is that a reason to dislike
chameleons?"
It is in the nature of x^y to have interesting behavior at the
origin. Apparently that upsets you in some way. I am unable to
comment further.
--
Dave Seaman
a...@seaman.cc.purdue.edu
migl...@cartan.math.nd.edu
> In article <1991Jan3.1...@news.nd.edu>
migl...@cartan.math.nd.edu writes:
> >Incidently, this is the way Lang does it. For a multiplicative group G
he
> >defines x^0 = e, x^n = x*...*x n times (for n positive) and x^(-n) =
> >(x^(-1))^n. You'll note that this doesn't have anything to do with 0^0
> >yet. I just want to point out that this is the standard definition.
>
> Then, a little later...
>
> >Sure, just as 0/0 = 1 is a special case of "x/x = 1 for all real x".
It's
> >your proposition which is in question, not your instantiation.
>
> Look again at what you just wrote above. You just agreed (quoting Lang)
that
> x^0=e for all x. So why are you saying that the proposition is in
question?
> You agreed that the general proposition is correct, and you agreed that
the
> instantiation is correct. What is your point?
Because G is a multiplicative group so x = 0 isn't allowed in that
statement (just as it isn't allowed in the x^(-n) part). You're doing
something non-trivial by including it.
>
> It is in the nature of x^y to have interesting behavior at the
> origin. Apparently that upsets you in some way. I am unable to
> comment further.
In the words of the warden in Cool Hand Luke (?), what we have here is an
inability to communicate. I'm sorry. In any case your original argument
is interesting so thanks for that.
Dave Seaman
>> Look again at what you just wrote above. You just agreed (quoting Lang)
>that
>> x^0=e for all x. So why are you saying that the proposition is in
>question?
>> You agreed that the general proposition is correct, and you agreed that
>the
>> instantiation is correct. What is your point?
>
>Because G is a multiplicative group so x = 0 isn't allowed in that
>statement (just as it isn't allowed in the x^(-n) part). You're doing
>something non-trivial by including it.
Look in the chapter on monoids in Lang. He says (and justifies) the statement
that m^0 = e for every m in a monoid M, based on the empty product argument.
You don't need a group. You don't need inverses. All you need is an empty
product. This is what I have been quoting all along. There is nothing special
about 0 in this argument other than the fact that it does not (except in the
trivial ring) have an inverse. If you want the argument to apply specifically
to 0, all you need is to take a ring with unity, ignore the addition operation,
and notice that you now have a monoid with a 0 element.
Incidentally, I'm sorry if I offended you earlier by assuming that you were
unfamiliar with basic algebra, but I had a number of reasons for drawing that
conclusion, based on your posting of yesterday. I will not discuss them on the
net, but you are welcome to send me mail. It was an honest mistake on my
part, and I apologize.
--
Dave Seaman
a...@seaman.cc.purdue.edu
Robert Strawderman
I believe that this discussion should be ended or moved to a
special newsgroup (like, sci.math.who.gives.a.FAQ). I don't
know how others feel, but I'm getting really tired of it...
Herb Brown
Are you certain about your spelling? It looks wrong!
Freek Wiedijk
>Doesn't it seem redundant to you to have two different notations for
>
> g(a) and lim g(x)
> x -> a
>
>if they are always supposed to mean the same thing?
He must be a Mathematica user. Mathematica evaluates:
Limit[g[x], x->a]
to:
g[a]
for an unspecified function g.
Greetings,
Freek "the Pistol Major" Wiedijk E-mail: fr...@fwi.uva.nl
#P:+/ = #+/P?*+/ = i<<*+/P?*+/ = +/i<<**P?*+/ = +/(i<<*P?)*+/ = +/+/(i<<*P?)**
Daniel E. Platt
] In article <3...@ndla.UUCP> pl...@ndla.UUCP (Daniel E. Platt) writes:
] >The problem with 0^0 = 1 is that it depends strongly on where you start, and
] >depends on what you're willing to bring to the problem (what context).
]
]True. I have stated before that there is more than one way to define
]exponentiation on the integers, which can then be extended to the rationals and
]the reals.
]
]>Algebraicly, x^0 is x/x. If x = 0, you're asking what number times 0 yields 0.
]>ALL numbers times 0 yield 0, so, x/x = 1 for x != 0, else its undefined.
]>In this context, strictly speaking, 0^0 is undefined.
]
] First of all, x^0 is x/x ONLY ON THE ASSUMPTION THAT x != 0. Therefore, you
] can't use that equation to draw conclusions about 0^0. As I have pointed out
I can use any definition I want to talk about 0^0, including x^0 = x/x...
from this, it follows that 0^0 is undefined. There's nothing wrong with
this assertion... except that it gives an answer you don't seem to want to
see or accept.
] before, 0^0 is an empty product, which is equal to 1. Another example of an
] empty product is 0! (zero factorial), which is 1.
0! is not an example of an empty product. It is just a number... the results
of evaluating f(n) = n! for n = 0. You can extend the idea of empty products
so that 0!=1 is an example if you want to, but this is a specific justification
in terms of cleaning up some definitions to make a homomorphism.
]
] >Analysis is another way people approach this problem. Depending on how you
] >take limits you'll see the desired results. For example:
]
] I think the problem is not "how you take limits" but how you apply the results.
]
] >[...] An example of something that DOES NOT work is
] >
] > lim_{x->0+} 0^x = 0.
]
] I don't know what you mean when you say that this example does not work.
] Your computation is correct, and therefore it "works." Perhaps it's just that
It does not work because it does not yield '1' for an answer. As such, it
proves the fact that x^y is at least discontinuous at x = 0. The issue is
whether x^y is even defined at x=y=0 and if so, what value does it have.
]you are trying to draw an unwarranted conclusion by assuming continuity at the
]origin? The problem lies in your logic, not in the definition of
]exponentiation.
Why should this be any more warrented or unwarrented than other starting
points? This result is just as valid as any following from other arguments.
]
]>
]>In particular, in the plane:
]>
]> lim_{x,y -> 0+} x^y = ?
]>
]>depends on how the limit is evaluated.
]
] No. The limit you indicate does not exist. If you get a different result,
] then your method of evaluating the limit is erroneous.
The limit exists for particular trajectories y=y(x) (or vice-versa). There
is no limit for all x, y. That's what I meant when I said 'It depends on
how the limit is evaluated.'
]
] >Lastly, combinatorics would suggest for consistancy's sake that, to preserve
] >the consistancy of a particular formula, that 0^0 = 1... but that is just a
] >special case of a special formula (in my book at least...).
]
] What do you mean by special case? Technically, 1^1, 3^2, and 5^7 are all
] "special cases" of the definition for x^y. Why should 0^0 be any more or less
] special than all the other special cases? And why is the formula any more
] special than, say, the product rule for derivatives?
0^0 is a special case different from any of the other numerical examples
x^y where x != 0. If it weren't a special case in at least some ways, this
discussion wouldn't be happening.
]
] >The original question was WHY are there so many disparate answers... the
] >answer to why there are so many is that there are so many ways that the
] >question can be answered, and many of them depend on particular contexts.
]
] It is true that the question can be answered in more than one way, but no one
] has yet shown how to answer the question so that a result other than 0^0 = 1
] is obtained. That is exactly the point of this discussion. When 0^0 = 1
] arises in more than one way from possible definitions, why is there so much
] resistance to adopting it? As I said at the beginning, "give me a reason."
]
Ok... 0^0 = 0 since lim_{x->0+} 0^x = 0.
Ok... 0^0 is undefined. It has no value. It doesn't make sense to give it
a value no way, no how because you cannot divide by 0. (0 is a special
element in any multiplicative ring since multiplication is defined in terms
of addition through the distributive axiom; it has no multiplicative inverse).
I've given reasons for both of these positions above. Take it or leave it.
You've asked why other positions would be stated. I'm not arguing merrits,
I'm just pointing out that these other positions exist because of the above
arguments.
If you want to get to the merrits of the case, 0^0 is a special case which
can only be handled through extensions to other fields, and then, mostly
as a definition to make algebraic homomorphisms possible, or to make a
summation easier:
f(x) = f(0) + f'(0)x + f''(0)x^2/2! + ...
is easier to write as a summation if x^0 = 1 for ALL x, for example (same
goes for 0! = 1; it makes the above summable). Because it is a special case,
and because it makes so many things more convenient, it is handled as a
definition rather than as a theorem.
While there are lots of reasons to take it as a special case; the people who
will happily take it on the one hand will often just as readily indicate that
they're treating it as a special case on the other hand. The reason is that
the situation is ambiguous. Most people are willing to acknowlege this state
of affairs and live with it. It doesn't fix anything for them to MAKE 0^0 = 1
because they don't really feel that anything was broken to begin with.
This is the way math has been handled. Of course, math has a lot of freedom
to play around with in terms of definitions, postulates, &c. This is what
makes math so much fun. If you don't like the fact that many people do
the above, you can write your own math books.
Dan
Dave Seaman
[Much drivel that has been refuted many times over already.]
In case you haven't noticed, the net readership is saying quite clearly that
this discussion has gone on long enough. If you still don't understand why
0^0 = 1, then
(1) read all the postings on the subject,
(2) if you still don't understand, don't post. Send me mail at
a...@seaman.cc.purdue.edu. I will explain it to you.
The same goes for everyone else who is reading this.
I also would like to add an appeal that the FAQ posting not be repeated until
the discussion of 0^0 is fixed, and that it not be posted more than once a
month in any event.
--
Dave Seaman
a...@seaman.cc.purdue.edu
Vidhyanath K. Rao
migl...@cartan.math.nd.edu writes:
>Are you saying that (b) doesn't give 0/0 = 1? There's only one way to
>write the empty set as a disjoint union of distinct sets of 0 elements,
>namely A = A.
To define addition one uses disjoint unions, not unions of ``distinct'' sets.
The definitions of disjoint union that I have seen are equivalent to this:
The disjoint union of $X$ and $Y$ is
$X \times \{0\} \union Y \times \{1\} \subset (X \union Y)\times \{0, 1\}$.
So the disjoint union of the empty set with itself is still empty.
All this flap about $0^$ seems to center around one question:
Is 0 as natural as 1, 2, 3?
If you say yes, as Dave Seaman seems to, you define exponentials first as
$\Naturals \times \Naturals \to \Naturals$ and then extend it. Here,
$0^0 = 1$.
If you say no, as lot people said till recently, and as many people on the net
still seem to believe, you define exponents first as $\Integers \times
\{1, 2, 3, \dots\} \to \Integers$ and then extend it. With this approach,
$0^0$ is undefined.
Before you say something about $0^0$, ask yourself, ``Is 0 as natural as 1, 2,
3?''
I think that knowing the historical developement of ideas is a ``good thing''.
If you think that this clutters up your mind, IMHO, you need a bigger mind.
--
Vidhyanth Rao It is the man, not the method, that solves
function.mps.ohio-state.edu the problem. - Henri Poincare
(614)-366-9341 [as paraphrased by E. T. Bell]
Robert D. Silverman
>In article <BROMAN.90D...@schroeder.nosc.mil> bro...@nosc.mil writes:
>>a...@seaman.cc.purdue.edu wrote:
>>> I have posted two
>>> good reasons for 0^0 = 1, and no one has yet posted a reason for any other
>>> interpretation.
>>
>>"0^0 is indeterminate" means only this: viewing exponent(x,y) as a
>>function of two real variables, a value for exponent(0,0) cannot be
>>defined such that exponent(x,y) would be continuous at x=0, y=0.
>
>[much about limits deleted]
>
>I agree absolutely with every word you said, which is nothing more than a
>lengthy rephrasing of what I said in an earlier posting: "The function f(x,y)
>= x^y has an essential discontinuity at the origin."
>
>So what is your point? I notice you never once addressed the central issue:
>What is the VALUE of 0^0? Limits are irrelevant, as I keep explaining.
One cannot separate the concept of the VALUE of an expression from the
concept of a limit.
You keep asserting that limits are irrelevent, but it is a mere assertion.
The fact that you assert something does not make it true.
The VALUE of a function f(x,y) at the point (x0, y0) IS lim x --> x0,
y --> y0 f(x,y). If the function is essentially discontinuous at (x0,y0), then
the value of f(x0,y0) must remain indeterminate.
Defining 0^0 = 1 is a definition that is useful in some circumstances.
However, this definition is not uniform throughout mathematics.
One can use your arguments to show that 0/0 = 1, yet it is generally
agreed that 0/0 is indeterminate. After all, x/x = 1, right?
--
Bob Silverman
#include <std.disclaimer>
Mitre Corporation, Bedford, MA 01730
"You can lead a horse's ass to knowledge, but you can't make him think"
Robert D. Silverman
:
:>Q: What is 0^0 ?
:>
:>A: According to some references, 0^0 is an undefined form
:> other books give it as 0 and a few others say it is 1,
:> the problem is that there are some good reasons for
:> choosing any of the three values.
:
:If there are good reasons, why is no one able to give one? I have posted two
:good reasons for 0^0 = 1, and no one has yet posted a reason for any other
mathematicians associated with this newgroup perceive the futility of
trying to argue with fools whose minds are already made up. It's a
waste of time. Most of us just can't be bothered.
Ilan Vardi
>
>It's not a question of being unable. It's just that the professional
>mathematicians associated with this newgroup perceive the futility of
>trying to argue with fools whose minds are already made up. It's a
Actually, that one of the main reasons I read this newsgroup. For
example, don't most people go to conferences to find out that no one
else has done anything nontrivial either? Just sitting around and
reading Selberg's collected works rapidly leads to terminal
depression.
Andrew Wheadon
Let the discussion continue until people do not want to continue discussing
it, till now I find it quite interresting, and I want to know wether they
will ever come to a conclusion
Bye Andrew and...@acwbust.incom.de
Robert D. Silverman
:>I don't know if analyzing 0^0 using limits has been presented, since I've just
:>joined this thread.
:
:Limits are not relevant to the discussion. It would be circular reasoning.
:
:The steps in definining exponentiation are:
:
: 1. Define exponentiation on the integers. There is more than one way
: to do this, as I pointed out in a posting earlier today.
:
: 2. Extend the definition to the rationals.
:
: 3. Extend the definition to the reals.
:
: 4. Extend the definition to the complex numbers (but this is outside
: the scope of the present discussion).
:
:The result 0^0 = 1 is obtained in step 1, as I have pointed out before.
:Limits are not even available as a tool until you get to step 3. Step 3 does
:nothing to change the results already established in steps 1 and 2. It merely
:extends the domain.
:
:Dave
:--
:Dave Seaman
:a...@seaman.cc.purdue.edu
But step 4 DOES. The value of x^y as x-->0 and y-->0 is path dependent.
Path dependence is an issue over C, but not over R.
S. K. Bhaskar
>
> One cannot separate the concept of the VALUE of an expression from the
> concept of a limit.
I am not quite sure what you mean by this. The value of an arbitrary function
( expression) at a point is independent of its limit value at that point.
>
> The VALUE of a function f(x,y) at the point (x0, y0) IS lim x --> x0,
> y --> y0 f(x,y).
Not for arbitrary functions. If f is continuous at (x0,y0), then
lim (x,y) -> (x0,y0) f(x,y) = f(x0,y0).
> If the function is essentially discontinuous at (x0,y0), then
> the value of f(x0,y0) must remain indeterminate.
Why ? f(x0,y0) could have any value assigned to it. If the value assigned
happens to make it discontinuous at (x0,y0), then, too bad, but we still
have a function with a well-defined value at (x0,y0).
>
>
> --
> Bob Silverman
> #include <std.disclaimer>
> Mitre Corporation, Bedford, MA 01730
> "You can lead a horse's ass to knowledge, but you can't make him think"
You seem to be implicitly assuming that the function f is continuous
throughout its domain. Perhaps you did state this somewhere and I missed it.