I read on Wikipedia that 0^0 = 1 is a definition. Is that also true for a^0=1, or does it follow from the a^b definition?
it’s also a definition. A^2 is AxA; A^1 is A; but A^0? it is a number but counted 0 times… by definition it is settled to 1.
0^0 could be 1 by the first definition, or undefined, or =0 (since 0^n =0), so settled to 1 for coherence.
Rather than employ a lot of math symbols that people may or may not understand, I prefer to think of the the meaning like this: On the average, when a person contracts a given disease, how many many other people is that person going to infect with the same disease?
For example, according to the source I accessed, Small Pox has an R-naught of “3” because an infected person, on average, will infect 3 other people. Polio was rated as “4” or higher.
Let A be a set of a elements (possibly a = 0, or a = ℵ₀ , doesn’t matter). Consider the empty set ∅ with zero elements. There is a single function ∅ → A, namely the empty function, so in that sense a⁰ = 1.
If you mean like in analysis, if you write, e.g., a^b = e^{b\log a} you can see there will be some issues near 0 so in any case you had better be clear what you mean if you are going to write something like 0^0 to indicate a limit. Of course, you are probably not going to write 0^0 = 2, regardless of the context!
But how does this work with the OP?
It would seem that A^0 means something like “how many people would A people infect if R0 is 0?” Which, I guess, one would logically assume to be 0. But that’s not the definition of A^0.
Similarly, 0^0 would mea “How many people would 0 infected people infect if R0 is 0”. Which should, logically, be 0. But that’s not how 0^0 is defined.
Basically but A^0 and 0^0 don’t have “logical” answers based on what exponents mean. So mathematicians have defined them both to be 1 because it makes certain other properties of those operators make sense.
I think you are talking about something different. But we do need some way (and/or math symbols) to interpret these R0 values because it is defined as some sort of average, which does not tell you everything about the disease model.
The fact that a^0 = 1 follows from the general properties of powers. For any number a \neq 0 , we have a^n/a^m = a^{n-m}. This relationship doesn’t work if a = 0, since then we’d be dividing by zero, but it should work for any other number.
If we accept this as true, then it must be the case that (a^n)/(a^n) = a^{n-n} = a^0. But any number divided by itself is 1. So a^0 = 1 for any number a \neq 0.
Well, the word “definition” is used in the OP and, to me, “definition” = “what something is”. I think very few people (definitely including me) can look at all that math and have and view it as a definition even though it is great stuff, so my contribution was, in essence, “R-Naught for Dummies”. LOL
But what does R0 have to do with this thread? This is about pure math, not epidemiology.
To the OP, there are properties of powers that mean that, in order for the properties to be consistent, any number to the 0 power must equal 1. There are also properties of powers that mean that 0 to any power must equal 0. Clearly, these two properties conflict in the case of 0^0. The usual approach is to declare 0^0 an invalid expression, or undefined. But in practice, when it comes up, it’s more common for it to be reasonable for 0^0 to be interpreted as 1 than for it to be interpreted as 0: For instance, the limit of x^x as x approaches 0 along the real number line is 1. So some sources, instead of considering 0^0 invalid, instead define it as 1.
So the question is, is a^0 defined to be 1 when a \neq 0, or is that a fact derived from a more general definition of a^b? And the answer, I think, depends on which definition of a^b one is using. But if it is true by definition, that’s the only way to define it that would be consistent with how other, nonzero exponents work.
Was this where you read it?
As that Wikipedia article notes, 0^0 is either defined to be 1 or left undefined, depending on the context. And it then goes on to discuss some of the various contexts, and why it might be defined as 1 or left undefined in those contexts.
First off, the basic identity involving exponentials is x^{a+b}=x^a *x^b. This is obvious when a and b are positive integers and remains true when the exponential is extended to negative numbers, fractions, real numbers, etc. (even complex numbers). The formula requires that for x\neq 0, x^0=1. This result is not forced when x=0. The real problem is that the two variable function x^y is discontinuous at (0,0). It was mentioned above that if you come in along the x axis, the limit is 1. But if you come in along the y axis, the limit is 0. Along the line x=y, it is not hard to show it is also 1, incidentally. But there are other reasons why mathematicians will generally take as definition 0^0=1, which I will not go into. But it is really a definition or, as I would call it, a convention.
That is the most succinct answer.
There are two implicit questions in the original post. The first is what do we mean when we write 00? That is a question of notation and the answer is a convention.
The second is what is the behavior of ab when a and b are near 0? That is a question of limits and requires analysis.
This is how I explain it to my students. I teach high school so there is some handwaving but the blanks can easily be filled in by someone who passed a class in abstract algebra.
There are two “magic” numbers called the identity. For addition the identity is 0 meaning for any number n that n+0 = n. For multiplication, the identity is 1 meaning n * 1 = n. For now let’s just focus on multiplication. When we multiply, we can fill blanks in with a 1, so when you have to factor out x from x + xy, we can look at the first term as blank times x and filling in that blank we get 1x + xy. Since division is a form of multiplication and fractions mean division, we see this when reducing. For example reducing 2/12 we can write it as 2 / (2 * 6) and the twos cancel out and we are left with a blank over 6. Well since it is really multiplication what do we fill in the blank with? A 1.
(Yes I know people will tell me there are ways of doing that without leaving a blank. That’s not the point. I’m showing students that based on their prior experience we fill in empty terms with a 1 when multiplying is occurring)
Back to the OP and let’s work backwards. Exponents are a multiplication problem. 5^3 = 5x5x5; 5^2 = 5x5; 5^1 = 5; 5^0 = . That’s it, there are no 5s to multiply so we are left with a blank. And when we have a blank with multiplying we fill it in with . . . 1. Thus any number n^0 would by that logic be equal to 1.
But 0^0 is an odd one. There are a lot of good reasons it should be equal to 1 but if you try to prove it, you eventually end up with the problem that for any number n that n x 0 = 0 so we mathematicians just define 0^0 =1 when it is absolutely necessary to have a value.
a^b could be seen as the number of ways you can pack b items into a boxes, assuming a box can contain multiple items. Given that, it’s obvious that a^0=1 , since there’s just one way to pack no items into some number of boxes. It also implies that 0^0=1, though as stated, that is more definitional.
By the way, I think I might have been a bit too conservative in this statement. I thought that I had remembered that the limit of x^x as x approaches zero was undefined, if x was allowed to be nonreal. But I double-checked, and (1e-6 i)^(1e-6 i) is very close to 1+0i, as is (1e-6 + 1e-6 i)^(1e-6 + 1e-6i), so I think that the limit does actually work in the complex plane as well.
I am definitely in the camp that would prefer for 0^0 to be left undefined. If different limiting calculations give different answers depending on the direction of approach then there is no answer its the same reason we leave 0/0 undefined even though the limit of x/x =1 as x->0 on the number line. I can see that in certain contexts is might be useful to accept a particular limit so that you don’t have keep including it as a side case, but I’m not comfortable with just defining it to be so in general.
It’s not just about the direction of approach. If you take x^y and approach along literally any straight path other than x=0, you get 1. Actually, any polynomial path will also get you 1. But there are other, more complicated, curved paths you can take to make it 0, or any number in between, or even any real number.
And that is what bugs me.
f(a,b) =(a)^b is just about the most well behaved continuous analytic function on the real numbers that one could ask for on its domain, which includes a>=0 and all b with the exception of (0,0).
For all a and b in this domain
lim_{(x→a)}f(x,b)= lim_{(x→b)}f(a,y)=f(a,b).
But at (0,0)
lim_{(x→0)}f(x,0)=1
and
lim_{(y→0)}f(0,x)=0
There is no particular reason to say we like the first limit but not the second. Far better just exclude the problem point and leave all the function’s lovely properties intact rather than shoe horn in an arbitrary choice be left with a functional kluge that is not even continuous much less infinitely differentiable.
No, but 1 is also the limit of (x^2+x)^{x^3-2x}, and x^{sin(x)}, and sin(x)^x, and x^{cos(x)-1}, and x^{e^\frac{-1}{x}}, and a bunch of others. It’s sorta hard to find functions that evaluate to 0^0 at x=0 that don’t have a limit of 1 (aside from variations on 0^{f(x)}).
Ok what about
f(x)=-1/ln(x) → 0 as x → 0
x^{f(x)}= exp(-ln(x)/ln(x))=1/e
But none of this matters. If a function takes on multiple values at a point it’s not a nice well defined function regardless of how you got there.
look at the graph in the wikipedia article linked above. There is obviously something messed up going on at (0,0). For any value between zero and one we can find numbers x and y arbitrarily close to zero such that x^y equals that number. Rather than calling all the points along that big long line equal to 1, why not just admit that there is a problem?
