hard maths question

0^0 = underfined

０＾0 = 1

0^0=?

If the exponent is positive, the power of zero is zero: 0n = 0, where n > 0.

If the exponent is negative, the power of zero (0−n, where n > 0) is undefined, because division by zero is implied.

If the exponent is zero, the power of zero is one: 00 = 1.

REASONS

From the power-series point of view, identities such as



圖片參考：http://upload.wikimedia.org/math/7/3/0/730670659edddec7251adf4bd1ca5ac7.png

are not valid unless 00, which appears in the numerator of the first term of such a series, is 1.

A striking instance is the fact that the Poisson distribution with expectation 0 concentrates probability 1 at 0; which agrees with the formula for the probability mass function of the Poisson distribution only if 00 = 1.

The consistent point of view incorporating these aspects is to define



圖片參考：http://upload.wikimedia.org/math/7/2/6/726c7ad6c7ae73185c35a44345dfaeb0.png

Here is another article about the discussion of 0^0


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 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.

0^0 = 1