What is 0^0
According to some Calculus textbooks, 0^0 is an ``indeterminate
form''. When evaluating a limit of the form 0^0 , then you need to
know that limits of that form are called ``indeterminate forms'', and
that you need to use a special technique such as L'Hopital's rule to
evaluate them. Otherwise, 0^0 = 1 seems to be the most useful choice
for 0^0 . This convention allows us to extend definitions in different
areas of mathematics that otherwise would require treating 0 as a
special case. Notice that 0^0 is a discontinuity of the function x^y .
This means that depending on the context where 0^0 occurs, you might
wish to substitute it with 1, indeterminate or undefined/nonexistent.
Some people feel that giving a value to a function with an essential
discontinuity at a point, such as x^y at (0,0) , is an inelegant patch
and should not be done. Others point out correctly that in
mathematics, usefulness and consistency are very important, and that
under these parameters 0^0 = 1 is the natural choice.
The following is a list of reasons why 0^0 should be 1.
Rotando &Korn show that if f and g are real functions that vanish at
the origin and are analytic at 0 (infinitely differentiable is not
sufficient), then f(x)^g(x) approaches 1 as x approaches 0 from the
right.
From Concrete Mathematics p.162 (R. Graham, D. Knuth, O. Patashnik):
Some textbooks leave the quantity 0^0 undefined, because the
functions x^0 and 0^x have different limiting values when x
decreases to 0. But this is a mistake. We must define x^0=1 for all
x , if the binomial theorem is to be valid when x = 0 , y = 0 ,
and/or x = -y . The theorem is too important to be arbitrarily
restricted! By contrast, the function 0^x is quite unimportant.
Published by Addison-Wesley, 2nd printing Dec, 1988.
As a rule of thumb, one can say that 0^0 = 1 , but 0.0^(0.0) is
undefined, meaning that when approaching from a different direction
there is no clearly predetermined value to assign to 0.0^(0.0) ; but
Kahan has argued that 0.0^(0.0) should be 1, because if f(x), g(x) -->
0 as x approaches some limit, and f(x) and g(x) are analytic
functions, then f(x)^g(x) --> 1 .
The discussion on 0^0 is very old, Euler argues for 0^0 = 1 since a^0
= 1 for a != 0 . The controversy raged throughout the nineteenth
century, but was mainly conducted in the pages of the lesser journals:
Grunert's Archiv and Schlomilch's Zeitshrift. Consensus has recently
been built around setting the value of 0^0 = 1 .
On a discussion of the use of the function 0^(0^x) by an Italian
mathematician named Guglielmo Libri.
[T]he paper [33] did produce several ripples in mathematical waters
when it originally appeared, because it stirred up a controversy
about whether 0^0 is defined. Most mathematicians agreed that 0^0 =
1 , but Cauchy [5, page 70] had listed 0^0 together with other
expressions like 0/0 and oo - oo in a table of undefined forms.
Libri's justification for the equation 0^0 = 1 was far from
convincing, and a commentator who signed his name simply ``S'' rose
to the attack [45]. August Mvbius [36] defended Libri, by presenting
his former professor's reason for believing that 0^0 = 1 (basically
a proof that lim_(x --> 0+) x^x = 1 ). Mvbius also went further and
presented a supposed proof that lim_(x --> 0+) f(x)^(g(x)) whenever
lim_(x --> 0+) f(x) = lim_(x --> 0+) g(x) = 0 . Of course ``S'' then
asked [3] whether Mvbius knew about functions such as f(x) =
e^(-1/x) and g(x) = x . (And paper [36] was quietly omitted from the
historical record when the collected words of Mvbius were ultimately
published.) The debate stopped there, apparently with the conclusion
that 0^0 should be undefined.
But no, no, ten thousand times no! Anybody who wants the binomial
theorem (x + y)^n = sum_(k = 0)^n (n\choose k) x^k y^(n - k) to hold
for at least one nonnegative integer n must believe that 0^0 = 1 ,
for we can plug in x = 0 and y = 1 to get 1 on the left and 0^0 on
the right.
The number of mappings from the empty set to the empty set is 0^0 .
It has to be 1.
On the other hand, Cauchy had good reason to consider 0^0 as an
undefined limiting form, in the sense that the limiting value of
f(x)^(g(x)) is not known a priori when f(x) and g(x) approach 0
independently. In this much stronger sense, the value of 0^0 is less
defined than, say, the value of 0 + 0 . Both Cauchy and Libri were
right, but Libri and his defenders did not understand why truth was
on their side.
