Why Does 0^0 = 1 and Not Undefined?
Date: 11/30/2007 at 15:31:42 From: Jesse Subject: 0^0 Your proof for why x^0 = 1 uses a law which breaks down at x = 0. Then in your definition for 0^0 you side significantly in favor of 0^0 = 1 based on your rule for x^0 = 1 (which was based on a law that breaks down at 0). Based on what I've read I would side in favor of undefined. Are there any more conclusive reasons for siding with 0^0 = 1?
Date: 11/30/2007 at 23:10:49 From: Doctor Peterson Subject: Re: 0^0 Hi, Jesse. I presume you are referring to our FAQs: N to 0 power http://mathforum.org/dr.math/faq/faq.number.to.0power.html 0 to 0 power http://mathforum.org/dr.math/faq/faq.0.to.0.power.html You can also find a restatement of the 0^0 issue here: Proof That 0/0 = 1 Based on x^0 Equaling 1? http://www.mathforum.org/library/drmath/view/69917.html Ultimately, the answer really depends on your context, as both 0^0 discussions above recognize to different extents. It HAS to be taken as an indeterminate form in calculus, because different limits that reduce to 0^0 have different values; we can't just say, "Oh, 0^0 is defined as 1, so that's the limit". But in many specific formulas or types of equations, taking 0^0 as 1 is necessary in order to write the formula without exceptions. For me, the clincher was when I realized that I've always talked about the constant term in a polynomial as the zero degree term, and yet if you write ax^2 + bx + c = ax^2 + bx^1 + cx^0 and don't take 0^0 as 1, you've changed the polynomial, which is defined for all x, into something that is undefined for x=0! I've never intended to do that, and never even thought about it, until a student pointed it out. Now I'm a believer: 0^0 not only must be, but simply IS, taken as 1 in many cases in ordinary algebra. But that doesn't mean we can automatically assume that it makes sense to define it that way in any context we come across. We have to keep our eyes open, and determine whether a new context is one in which this definition fits. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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