oops wrote:
yes, good work juaz and himalayan!
the answer C, although correct for almost all cases, it fails when M and N are both 0, bcoz 0^0 is undefined and the result is NOT an even integer.
since u both found it together, i suggest you split the extra credit
I agree that 0^0 could be any value (odd, even) , but I read smwhere that 0^0 = 1 is "widely accepted" now for practical purposes.
Try this on calculator and you'll get 1.
Ok...From Matforum:
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 0^x and x^0 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 of 0^0 is very old. Euler argues for 0^0 = 1 since a^0 = 1 for a not equal to 0 .
Consensus has recently been built around setting the value of 0^0 = 1 .