We had this issue come up in calc while discussing exponential functions and couldn't come up with a clear answer. Any one know anything useful.
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We had this issue come up in calc while discussing exponential functions and couldn't come up with a clear answer. Any one know anything useful.
My calculator (TI-86) gave me a domain error. I have class in about 30 mins and I'll ask the profs what they think.
So for now I'm saying undefined
I know that that is what the 83 says, but apparantly there are reasons for it having certain values, but that was all my teacher could remember from some class he took ages ago.
I think the confusion comes due to conflicting rules
0^x = 0
x^0 = 1
so 0^0 = would equal 0 and 1.
I'll ask though. The prof that has the class before mine knows a great deal and my prof just got out of college a few years ago so she might still remember :)
google does miracles...ever hear of it?
http://www.faqs.org/faqs/sci-math-fa...lnumbers/0to0/
That is really interesting, so apparantly they can't make up their mind about it.
Off hand, I think my calculus textbook said something about the function 0^x approaching 1 as x approaches 0. However, 0^0 itself is considered to be undefined.
I asked and the answer I got is that it is generally considered to be undefined but some text do define it as 1 or 0.
Edit: I forgot they mention that when the text's do define it they do it by convention in the same manner that 0! is defined to be 1. Ok time to get rid of this headache :)
out put:Code:#include <iostream>
#include <cmath>
using namespace std;
int main()
{
cout << pow(0.0,0.0);
cin.get();
return 0;
}
1
My university "textbook" (a collection of sheets actually) leaves it undefined. The whole thing is approached in the context of limits however. According to my textbook, the limit of something that results in an indeterminate form must be determined by looking at the functions involved. Usually the rule of l'Hospital is used.
There are some nice examples of how the limit of another indeterminate form, 0/0, depends on the functions involved.
Windows calculator says 0^0 = 1.
I think the Windows calculator might well use the pow function internally, which obiously uses the optimization of testing the second parameter for 0 and automaticaly returning 1 in case it is.
Using a C++ program or the Windows calculator would be unreliable in this case, since neither is an authority on such a definition, but is merely implementation dependent.
It doesn't exists, since the limit differs depending on which axis you move along.
lim (x -> 0) x^0 = 1
lim (y -> 0) 0^y = 0
I was following along fine until i hit this post.....thnx magosQuote:
Originally posted by Magos
It doesn't exists, since the limit differs depending on which axis you move along.
lim (x -> 0) x^0 = 1
lim (y -> 0) 0^y = 0