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0^0

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Re: 0^0

#181

Post by lotharbot »

einhverfr wrote: post #164
ndeine wrote: post #155

Earlier quoted context omitted.

The relationship f(x)=x/x is only defined for x /= 0 and thus is not equivalent to 1. We can see this in [a] when we ask for the domain of the function, and thus we can redefine some function f' as a piecewise function which is defined to have f'(0)=1, but in proofs we must thus make sure to first prove that using f' as a substitute for f does not affect our result. In one of the below posts we have the suggestion >…

> The relationship f(x)=x/x is only defined for x /= 0 and thus is not equivalent to 1. If that's the case then: x * f(x) = x cannot be equivalent to f(x) = x/x which breaks algebra in pretty fundamental ways (since the former would certainly be defined for 0 but the latter would not). The second major problem is that it also breaks calculus. Let's start with a straight line: f(x) = 2x. Now let's take the first deriv…

> "x * f(x) = x cannot be equivalent to f(x) = x/x"

Sure it's equivalent, over a domain not including x=0. This does not break algebra any more than, say, restricting the domain of the square root (when working in the reals) to non-negative numbers. We work in restricted domains in mathematics all the time.

> " f'(x) = 2x/x"

f'(x) = lim (h->0) [2(x+h)-2x]/h. Since h is approaching (and therefore not equal to) zero, there is no problem. Any appearance of 0/0 in the problem is a result of an attempted (but unsuccessful, that is, indeterminate) evaluation -- it's not actually 0/0, it's 2h/h where h is close to but not equal to 0.

We don't need to define 0/0=1 in order to have either algebra or calculus work. We choose to define 0/0=1 in certain circumstances which make certain calculations go more smoothly, and we choose not to define 0/0 in other circumstances where it's either unnecessary or potentially misleading.

Re: 0^0

#182

Post by dagw »

LanceH wrote: post #161
noonespecial wrote: post #106

Earlier quoted context omitted.

The utility of math is that you can then do a bunch of calculations with donkapples and at the end, get a meaningful real-world answer.

Well meaningful as soon as someone finds a donkapple.

Not at all. Complex numbers are used to solve real world engineering problems, despite the fact that you cannot "find" any complex numbers in nature.

Re: 0^0

#184

Post by ivansavz »

vlasev wrote: post #48
forrestthewoods wrote: post #31

Earlier quoted context omitted.

I love Wikipedia. It's amazing. It makes the world a better place. I'm a pretty decent programmer. I do video games so I do lots of 3d math. I'd say I'm decent at that as well. I hate Wikipedia for math. Absolutely hate it. Unless you are a mathematician by trade Wikipedia is damn near useless for learning new math concepts. I don't even bother checking it anymore.

I had an idea for a couple of years now of creating a wikipedia-style mathematics textbook that will be crowd-sourced, standardized and cover all of math in a way that's accessible to learn from on your own. It would have a kind of a zoom function where you can expand details on explanations and calculations to a depth that you prefer. Ideally this kind of thing would start off with basic math and get progressively f…

That is not a bad idea at all.,, It will be a lot of work but it could be done. Crowd sourcing might lead to the wikipedia-style schizophrenic-author voice of writing, though. See [1] for a prev discussion.

Interestingly, if the book is organized so it is aligned to the common core math standard[2], it would make a killing in the US market since current textbooks are so bad.

[1] https://news.ycombinator.com/item?id=7456397 [2] http://www.corestandards.org/Math/

Re: 0^0

#185

Post by trentmb »

SamReidHughes2 wrote: post #173
rickhanlonii wrote: post #141

Earlier quoted context omitted.

I agree with the point you're making, but I want to be a little pedantic: It's true that addition is a definition, but 1+1=2 is not--it logically follows from the definition of addition.

It's a definition of what "2" means.

I was always under the impression that 2 is shorthand for s(s(0)).

Re: 0^0

#186

Post by sold »

stephencanon wrote: post #78

The real problem here is that x^y is a single shorthand which refers to a few fundamentally different mathematical concepts (which happen to have significant overlap with each other). First, it refers to a function f:C x N --> C, defined in terms of repeated multiplication. f(x,0) is 1 for all x != 0, and so we adopt the convention that f(0,0) is also 1. But it also refers to a function g:C x C --> C, defined as g(x,…

I'd like to add x^y is also used other contexts - e.g. for cardinal numbers and ordinal numbers.

Re: 0^0

#187

Post by rapala »

chacham15 wrote: post #143
mathgrad wrote: post #137

Earlier quoted context omitted.

I am a mathematician I don't need examples. What is your formal definition of a mapping. In mathematic it is normally used as a synonym for a morphism in a given category and in the category of sets this would be a function. So in our context a mapping is a function.

> I am a mathematician I don't need examples. I gave the example not to teach you, but so that we could get away from terminology and get to meaning. Look at the question I posed: "how many arrows would it take pointing from the student to the teacher to illustrate every possible assignment?" So, if you have two students Alice and Eve, and two teachers Bob and Carl, there are four arrows: Alice->Bob, Alice->Carl, Eve…

The number of arrows is |students x teachers|, every student paired with every teacher. From this you get a^0 = 0, but it also follows that a^b = b^a. This defines multiplication, not exponentiation. The problem is that you cannot illustrate the assigment where Eve is not paired with Bob using an arrow, but only by not using a one.

Re: 0^0

#188

Post by Perseids »

_nedR wrote: post #178
rntz wrote: post #105

Earlier quoted context omitted.

In a certain sense, "1+1 is 2" is also merely a definition, in the same sense that "0^0 is 1" is a definition. Addition can be formally defined in mathematics; we habitually omit this definition because it is tedious, and because addition is such an intuitive operation that we do not require a definition in order to reason about it. Much as the question "what if the parallel axiom didn't hold?" leads to alternative g…

Its funny how quickly the argument devolved into an argument over how (or even whether) 1 + 1 = 2

It is not a devolution. The discussion whether 1+1=2 is a definition, an axiom, or a theorem is very similar to the discussion of "0^0=1". In each case we are looking at different way to construct a useful mathematic out of different sets of definitions and axioms. Regarding 0^0 there is no clear winner, as - like the blog shows - there are many incompatible advantages one would get from 0^0:=1 or 0^0:=0. With "1+1=2" there is no controversy, as the many different definitions or axioms essentially result in the same nomenclature.

But as an example of a rather unintuitive result regarding "1+1" look at the field F_2 (or GF(2)). It has two elements, 0 and 1. So when I'm writing "1" or "0" I am referring to these elements. Nonetheless with a typical definition used in many branches of mathematics I define 2:=1+1, 3:=1+1+1,… In F_2 the addition operation is defined as 0+0=0, 0+1=1, 1+0=1, and… 1+1=0. I thus get 0=1+1=2, 1=(1+1)+1=3, 0=((1+1)+1)+1)=4. Note that this is essentially calculation mod 2. But whereas outside of algebra this is often seen as a different set of equalities ("1+1≡0 mod 2") in algebra F_2 is just another field alongside of the rational, the real and the imaginary numbers.

Re: 0^0

#189

Post by einhverfr »

lotharbot wrote: post #181
einhverfr wrote: post #164

Earlier quoted context omitted.

> The relationship f(x)=x/x is only defined for x /= 0 and thus is not equivalent to 1. If that's the case then: x * f(x) = x cannot be equivalent to f(x) = x/x which breaks algebra in pretty fundamental ways (since the former would certainly be defined for 0 but the latter would not). The second major problem is that it also breaks calculus. Let's start with a straight line: f(x) = 2x. Now let's take the first deriv…

> "x * f(x) = x cannot be equivalent to f(x) = x/x" Sure it's equivalent, over a domain not including x=0. This does not break algebra any more than, say, restricting the domain of the square root (when working in the reals) to non-negative numbers. We work in restricted domains in mathematics all the time. > " f'(x) = 2x/x" f'(x) = lim (h->0) [2(x+h)-2x]/h. Since h is approaching (and therefore not equal to) zero, t…

> This does not break algebra any more than, say, restricting the domain of the square root (when working in the reals) to non-negative numbers.

Sure it does, because if that is the case, you restrict your domain when you divide by a variable expression. If you divide both sides by x-1, then you effectively rule out 1 from the domain.

That's the problem.

Now this is not the same as 0/0. The point is that 0/0 is only undefined when it persists after simplification and only because you can't define a relationship between the two zeros.

I.e. 0/0 is undefined because 2x/x, 52x/x, and x^2/x give you three different answers as x->0. That doesn't mean that every function which has not been reduced and can transiently evauate to 0/0 is treated as non-continuous.

Regarding derivatives, this highlights the problem because to solve the first derivative of a variable to a simple exponent, you multiply by the exponent and divide by the variable (x^2 becomes 2x, 2x becomes 2, and so forth). What this means is that you may be dealing with a limit but the limit defines a function, which is something like 2x^2/x for the derivative of x^2 and 2x/x for the second derivative.

Unless you allow simplification before determining whether the function is continuous, these things don't make sense. If you allow reduction first, then x/x^2 is undefined where x = 0, but x/x is not, because you can reduce it to 1 before applying any further logic. Both may appear to evaluate to 0/0 however.

There are a huge number of things that seem to break in algebra and calculus if one treats x/x as non-continuous and undefined. The simpler solution is to allow reduction to 1 before determining that it is undefined. (of course 52x/x would reduce to 52 instead, again showing why 0/0 is oversimplifying the problem).

Re: 0^0

#190

Post by Twisol »

einhverfr wrote: post #140
Twisol wrote: post #120

Earlier quoted context omitted.

I prefer thinking about 0^0 = 1 as an empty product ( https://en.wikipedia.org/wiki/Empty_product ), since it generalizes nicely to any operation with an identity element. That is, if you apply any operation zero times, the result is that operation's identity. It's interesting that the analogous empty sum, 0*1 = 0, is a complete non-issue.

But here's your problem: infinity * n = infinity, right? 0 * n = 0, right? 0 * infinity = ? Ok, this is relevant here particularly because: Lim 1/x as x -> 0 from the positive side is infinity, right? So 0 * that is..... lim 1/x as x -> from the negative side is negative infinity, right? So 0 * that is..... That's why 0/0 doesn't work as such. You don't know how 0 is derived or what it means. If we have x^2/x, and ta…

You are conflating the limiting behavior of a function with the value of a function. In (standard) analysis, there is no actual value called infinity - it's just used to describe how a function behaves arbitrarily close to a given value.

When you have a 0^0 limiting form (or 0/0, or 0 * infinity), the function's behavior is indeterminate. The "0" and "infinity" you're looking at aren't precisely 0 or infinity, but only arbitrarily close. The actual behavior of the function depends on the expressions that approach 0 and infinity, hence `x/(x+1)` and `x/e^x` having different limiting behaviors as x increases without bound.

But if you are literally considering the function at a specific point which produces a so-called indeterminate form, the answer is simpler. In some cases, the function is undefined (e.g. 1/x where x=0). In the case of 0^0, there is a precise value we can assign: 1. And this doesn't conflict with the 0^0 limiting form: the limit of a function at a point can be different from the actual value of the function at said point.

A quote of Knuth, taken from Wikipedia (http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_powe...):

> Knuth (1992) contends strongly that 0^0 "has to be 1", drawing a distinction between the value 0^0, which should equal 1 as advocated by Libri, and the limiting form 0^0 (an abbreviation for a limit of f(x)^g(x) where f(x), g(x) -> 0), which is necessarily an indeterminate form as listed by Cauchy: "Both Cauchy and Libri were right, but Libri and his defenders did not understand why truth was on their side."

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