# Quick maths question...

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Hi, I have a simple maths question 'cause I'm not gettin' something here. I was on wolfram alpha, and entered: ((2+3*x)^(a+b))^(a+b). Since it's based on Mathematica I will believe that the results I'm getting are correct. It shows an alternate form of the term I entered which is this: (3*x+2)^((a+b)^2). It says though that this only counts for real and positive numbers. I get why this wont do for imaginary numbers, but why only positive numbers? Cheers!

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if x was -2/3 and a,b are zero, then the expression would be 0^0 which is usually considered undefined.

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Oh well if you see it like that yea :P, to me thats 1. But I guess something like this would be true for an example case like so?

f(x) = x * x^-1.

I don't think there is some standard value set for 0/0 like there is for 0^0?

edit: I just found out that wolframalpha defines 0/0 as 1. I'm used to compilers saying 0^0 is 1, but I've never seen 0/0 as 1. I guess it's up to everyone to define these special cases for their purposes, or what would you guys say?

This also explains why my example from this post doesn't have any restrictions when evaluated by their engine.

[Edited by - BrickInTheWall on March 8, 2010 11:04:47 AM]

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I'm always wary of defining things like that, because if you think of it in terms of a function of 2 variables: f(x,y) = x^y or g(x,y) = x/y, the limiting value f(0,0) differs depending on which direction you approach from.

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Rutin
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