PureBasic Forums - English

thefool wrote: why is it an error? its NOT

Because it has not a REAL result, that's why it an error, same as 0/0 for example.
BTW, do you understand the meaning of REAL value?

Ohhhh, Niels Bohr was born in Denmark too. If he see this .... :roll:

DoubleDutch wrote:TheFool is correct, the answer is 1.

0/0 is solvable to one of two possible answers, its either +infinty or -infinity... - just beein discussing it with thefool.

btw 0/0 <> 0^0 'cause 0^0 is 1, not infinity!

-Anthony

0^0 is indeterminate (the answer depends on the problem). Think of the limit of 0^x as x->0+, then the answer is 0. Think of the limit of x^0 as x->0+, then the answer is 1. For convenience it is usually defined as 1 though (e.g. the Binomial theorem would need to be altered if it wasn't). It doesn't really matter what it is defined as though, as long as everyone knows what value it is. So in our case it is 1

DoubleDutch wrote:0/0 is solvable to one of two possible answers, its either +infinty or -infinity... - just beein discussing it with thefool.

btw 0/0 <> 0^0 'cause 0^0 is 1, not infinity!

-Anthony

Are you more than 11 y.o. ??
Sorry, but I can't believe what i am reading!!!

Are you more than 11 y.o. ??
Sorry, but I can't believe what i am reading!!!

huh?

Mr MAt, thank you.
I leave you here alone with them. Ouuuufff!!!

are we going to get rude now???

oh well you can say what you want. x/0 is ]-∞;∞[

are you telling me that x/0 is NOT that ???

okay as MrMat says: the answer is 1. it might not be but according to all
nature and human laws made it is defined as 1.


SO PSYCOPANTA: pb SHALL return 1 as everyone ELSE is doing. if it returned an erro it would be inacceptable..
i understand what you say, but mrmat points out that we use 1 as the answer.

Are you more than 11 y.o. ??

no. he is 7, i am about 5½

MrMat wrote:Think of the limit of 0^x as x->0+, then the answer is 0. Think of the limit of x^0 as x->0+, then the answer is 1.

Why do that??? Shouldn't be the logical way to think of the limit of x^x as x->0+ ?????

sorry read it wrong.. well but im not sure still! as its defined as anything ^ in 0 is 1.

so unless you can specify me a concrete evidence then i will still belive the anything ^ 0 is 1.

but ok you got a point..
but still my hardware evidence proves that DD and me is right..

Other than the times when we want it to be indeterminate, 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 would otherwise require treating 0 as a special case.

okay.. but according to this its correct that pb gives 1 as anwer..

thefool wrote:so unless you can specify me a concrete evidence then i will still belive the anything ^ 0 is 1.

Don't continue being a donkey, and ask yourself what about 0^anything.

And when you have the answer then think about anything^anything, and think about those "anything" is 0 (which is just my above question to MrMat)

Psychophanta wrote:

MrMat wrote:Think of the limit of 0^x as x->0+, then the answer is 0. Think of the limit of x^0 as x->0+, then the answer is 1.

Why do that??? Shouldn't be the logical way to think of the limit of x^x as x->0+ ?????

I was making the point that the answer depends on the question of what 0^0 actually means. If you want it to be the limit of x^x as x->0+ then 0^0 will be 1. The logical way would be to also consider the limit as x->0- and if all limits exist and are equal then you could say 0^0 = that value. At the end of the day 0^0 = 1 is a useful definition that fits in well with many formulas so thats why it may often be defined that way.

Let's ask Dr. Math...
http://mathforum.org/dr.math/faq/faq.0.to.0.power.html

...and MrMat, never forget that conveniences are always human interests