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Doesn't a simple application of l'hospital solve this?
That's addressed in the article under "cleverest student."
Technically, 0^0 is an indeterminate form and has no specific solution. Accurate but unhelpful. Practically, 0^0 highlights the issue that most of us don't have a good conceptual model for what exponents really do. How would you explain to a 10-year old why 3^0 = 1 beyond "it's necessary to make the algebra of powers work out". I use an "expand-o-tron" analogy http://betterexplained.com/articles/understanding-exponen…
I successfully managed to explain 3^0 to 10 year olds (as a recovering high school math teacher) as: 3^2 = 9 3^1 = 3 (divide 9 by 3) 3^0 = 1 (divide 3 by 3) 3^-1 = 1/3 (divide 1 by 3) etc This can logically be explained as n^0=1 for all real numbers. Unfortunately this doesn't really handle 0^0 but fortunately 10 year olds are rarely that difficult.
3^3 = 3*3*3*1 = 27
3^2 = 3*3*1 = 9
3^1 = 3*1 = 3
3^0 = 1 = 1
0^3 = 0*0*0*1 = 0
0^2 = 0*0*1 = 0
0^1 = 0*1 = 0
0^0 = 1 = 1
Doesn't a simple application of l'hospital solve this?
Earlier quoted context omitted.
I haven't seen any reason why -e^(i pi)=1 could be considered incorrect. It's consistent with the Taylor Series expansion of e^x. It's consistent with the view of complex exponentiation as rotation. I don't know of any particular problems that arise from taking -e^(i pi)=1. This is not the case with 0^0=1, which is inconsistent with many limits. That's why 0^0=1 is an agreed-upon convention sometimes . http://en.wiki…
This is not the case with 0^0=1, which is inconsistent with many limits. So what? It's only a problem if you want the exponentation function to be continuous, so you escape the problem by leaving it undefined. You could place similar unbased requirements on complex exponentiation to make it seem incorrect. For instance, real exponentiation always gives a positive value for positive base, while complex does not, so e^…
Even the original article noted that we don't choose the 0^0 convention because it's "correct", but because it's "nice" -- which is why we define it that way in the contexts where it makes sense to define it that way. If you're working with combinatorics, 0^0=1. If you're working with cardinal exponentiation, 0^0=1. If you're taking limits, or working in the hyperreals, or in certain other contexts, the convention doesn't apply. In some circumstances, 0^0 isn't even a valid statement -- like if you're working directly with the field axioms of R.
Recognize what context you're working in, and what assumptions or conventions apply in that context. That's just good mathematics.
Earlier quoted context omitted.
> Also, mathematicians agree in this , seriously. Go and ask one. Mathematician here; we do not. See http://math.stackexchange.com/questions/11150/zero-to-zero-p... . More precisely, as Arturo Magidin points out at http://math.stackexchange.com/questions/11150/zero-to-zero-p... , if we view exponentiation in the 'discrete setting', then $0^0$ must be $1$; whereas, if we view it in the continuous setting, there is sim…
But my point is, if 0^0 = 1 is _the_ answer in discrete setting, and it's _an_ answer in continuous setting, why don't we just agree that 0^0 = 1 and stop creating confusing situation where sometimes it's defined and sometimes it's not. 0^0 = 1 does not make calculus theorems more complicated to state or prove with modern language. It was a problem in XIX century, when mathematicians did not have a solid foundation w…
Because sometimes it's better not to. Sometimes it's inconsistent with our definitions.
Just like sometimes we agree that you can't divide by zero, and sometimes we agree that you can. Sometimes infinity is an actual value (say, in the extended reals), and sometimes it's just a symbol for "unbounded". Sometimes we agree that you can't take the square root of a negative number, and sometimes you can. Sometimes we use the axiom of choice, and sometimes we don't (and you can have an awful lot of fun either way!)
Mathematics is contextual. How various operations behave depends on which axioms and conventions are being used.
Earlier quoted context omitted.
But my point is, if 0^0 = 1 is _the_ answer in discrete setting, and it's _an_ answer in continuous setting, why don't we just agree that 0^0 = 1 and stop creating confusing situation where sometimes it's defined and sometimes it's not. 0^0 = 1 does not make calculus theorems more complicated to state or prove with modern language. It was a problem in XIX century, when mathematicians did not have a solid foundation w…
I'm a mathematician as well As am I, by training if not by profession. As is lotharbot. You're in a thread full of mathematicians. :) Which is what I would expect on this site, actually. I'm always timid making technical claims here unless I'm sure I'm correct; it seems to be a place frequented by arbitrarily large fish. To answer your question, if 0^0 = 1 is _the_ answer in discrete setting, and it's _an_ answer in…
From this point of view, the whole notion of "indeterminate form" makes just as little sense as distinguishing some arbitrary class of functions and calling them "elementary". Why are some points of discontinuity of some functions more special than other points of discontinuity of other functions? Why sin is more elementary than gamma? Historical heritage of confusion, I guess.
Sometimes you have to use another language to make sense of something.
I'll stick with the grade-school math approach, at least until I need to approach it differently. 4^2 = 2 fours multiplied = 4 * 4 = 16 divide by 4 - so you take away one of the 4s by division(canceling like terms like we do in grade school fraction math): 4^1 = 4*4/4 = 4 divide by 4 again 4^0 = 4*4/(4*4) = 1 divide by 4 again! 4^-1 = 4*4/(4*4*4) = 1/4 Now try it with 0: 0^2 = two zeros multiplied = 0*0 = 0 Divide by…
0^-1 is definitely infinity x^(-1) doesn't have an upper bound as x approaches zero.
Earlier quoted context omitted.
I'm a mathematician as well As am I, by training if not by profession. As is lotharbot. You're in a thread full of mathematicians. :) Which is what I would expect on this site, actually. I'm always timid making technical claims here unless I'm sure I'm correct; it seems to be a place frequented by arbitrarily large fish. To answer your question, if 0^0 = 1 is _the_ answer in discrete setting, and it's _an_ answer in…
Please, treat exponentiation just like every other function out there. I don't get the whole limit argument at all. Given any function f: R x R -> R, if it happens that a_n -> a, b_n -> b, but lim f(a_n, b_n) != f(a, b), people just say that f is not continuous in (a, b), and the case is over. However, if f happens to be exponentiation function, people instead argue that f should not be defined in (a, b), forgetting…
Similarly, if you're evaluating a limit and you get 0^0 you need to do more work. You can't just stop and say "oh, that's 1". It depends on what function you used to get there -- x^x will give you a different answer from ( e^(-1/x) )^x. Again, it has nothing to do with the continuity of exponentiation. The issue is whether the convention of 0^0=1 is correct in the specific part of mathematics you're working in.
The same argument can be made if you're working in the hyperreals, or if you're working with field axioms -- the convention 0^0 doesn't work in that context.
Please, by all means, use the convention 0^0=1 when it's appropriate. But understand that it's not always appropriate. Not every mathematician works in the particular subschool that you do; not every mathematician is going to find your convention appropriate.