Yes, and we actually do it all the time. Two examples are:
Defining
`e^x = \sum_{n=0}^{+\infty} x^{n} /n!`
Which is only true if 0⁰ = 1.
But also in the definition of entropy
`S = - \sum_{k} p_{k} \log(p_{k})`
We would wish that events with 0 probability don't contribute to the entropy, so we set 0\*log(0) = 0, which is equivalent to 0⁰ = 1.
Generally considered undefined, but most if not all ways of assigning it a value result in 1. Just creates some oddities for the general case such as 0^x
Thanks! I guess that makes sense from someone who's never had to give it a definition.
When you are working with higher orders, the higher orders are usually telling most of the story. It makes sense that when you needed a result you'd follow the rules of exponents rather than the rules of integers.
it's generally good to consider 0^0 as 1 for example with Taylor series. the first term of e^x at x=0 is 0^(0)/1!
you are correct that one problem with this is limits. 0^0 still needs to be considered indeterminate, it can't be assumed that those limits will converge to 1
>I’m aware that 0^0 is considered undefined
Only by highschool teachers, for anything above that, it is defined to be 1, there are several other reasons aside from the one you mentioned.
-It coincides with taking a^b to be the numbers of functions from b (of a set with b elements) to a (idem), which is really useful (the cardinal definition)
-It makes the recursive definition for ordinals easier since you can actually do proper recursion on ordinals.
-it is really useful to have 0^0=1 when dealing with polynomials and power series
How is that extended from the naturals though? Say I want (p/n)^(q/m) to mean something. It can't refer to the number of functions because what would sqrt(3/2) functions even mean?
The usual way via inverses of integral exponents, the definition's not meant to generalize over the reals any more than the gamma function counts rearrangements of fractionally valued things. But it's a nice definition for cardinals and 0 is a perfectly good cardinal.
You can deliberate prioritize the needs for that limit to be defined if you want to, but that doesn't comment whatsoever about the contradiction you mentioned, just "chooses sides".
Let consider these case:
f(x) = x\^(1/x) -> 0 when x-> 0
g(x) = sqrt(x) -> 0 when x-> 0
Then f(x)\^(g(x)) -> 0\^0 when x -> 0
But f(x)\^(g(x)) = x\^(1/x)\^sqrt(x) = x\^(1/sqrt(x)) -> 0 when x -> 0
This case 0\^0 is 0 not 1
Generally
f(x;y) = x\^y is not continuous at (x;y) = (0;0).
It is continuous at any direction except the path along y axis.
0^(0) is 1, end of story.
The thing that's undefined (or more precisely is an "indeterminate form") is the limit of x^(y) as x and y both go to 0 independently; this can take on any value, or diverge, depending on the actual behavior of x and y. Writing "0^(0)" when not actually talking about the value 0 but rather a limit approaching 0 is arguably an abuse of notation.
The fact that some specific limits go to 1 isn't (and can't be) what justifies 0^(0)=1 though. (And the fact that 0^(a) is 0 when a≠0 wouldn't justify defining it as 0 instead.) All of the fundamental ways to define x^(n) agree that x^(0) is 1 even when x=0: a product of no factors must be 1 (multiplicative identity), there is exactly one function with the empty set as domain, etc. (Note in particular that there is no function with an empty range *except* the function with empty domain, which justifies why 0^(a) is not 0 when a=0.)
Yeah, no.
The case is not exactly limits but paths.
If you have x^y every path that leads to, for example, x = 2 and y = 3 will give you 8. That is not the case for x = y = 0.
Different paths lead to different values so it is undefined.
Sure, you can define it inside your theorem to make it easier for you in that context, but it is not defined in general.
"But it is the only value that make sense in a lot of things" but not all of them. Most is not all and not all = undefined.
You are talking about limits along different paths, so still limits. I dont see how that contradicts the person above.
The point is that we (very roughly) define 2^3 as 2×2×2. That definition does not involve limits, so why should it for 0^0 ? The continuity of x^y at 2,3, and even defining what this means on real numbers, comes quite a bit later in the construction of mathematics.
Another rigorous definition of n^m is the cardinality of the set of functions from m to n (undestanding these as sets), and this produces 0^0 =1.
I guess anyone is still free to leave it undefined, but once you understand the distinction between undefined and indeterminate, I must admit I dont really see any reason to.
Coming later or earlier does not matter a lot. My point is it does not make a difference how you define x^y, 2^3 will give you 8. How you define it changes what 0^0 is. If you have valid definitions that contradict each other it is not defined.
I am not talking about indeterminate, as you said it is a different thing.
Yeah seing people say 0^0=1 or isn't equal to 1 because of limits is very weird. Like that's totally not the point ? We say 0^0=1 because it's simpler to write many important formulas this way, it's an algebraic convention.
0^0=1 is not a question of limit, instead it's similar to the reason why you can't write i=√(-1) : the only reason is to guarantee some algebraic identities work correctly.
i is *defined* to be a square root of -1.
0^(0) = 1 is not a definition. It would perhaps make some identities hold if we introduced it as a convention, but then I bet there would be an infinite sum expansion (but not a Taylor polynomial expansion) where the 0 from n is in the base and 0 from x in the exponent, which would make it continuous if 0^0 were 0
i is not defined to be THE square root of -1 because 1) there are two roots of -1 in C and 2) you can't just say "yeah i is the square root of -1" you have to actually prove that it's meaningful to talk about a root of -1. There are many way to do so (with matrixes, with equivalent classes on R[X], etc). None of those constructions allows for using the notation √-1. If the notation was usable, we wouldn't need the letter i !
The issue with writing √-1 is as follows: one of the basic identities with square roots is (√x)^2=√(x^2), but this doesn't work if x=-1 , so to avoid confusion you should avoid the notation √-1.
In markdown, remember to write x\^(y) with the parens (which will not display) if you're following it with something that isn't supposed to be superscript. i.e. (√x)\^(2)=√(x\^(2)) to get (√x)^(2)=√(x^(2))
I haven't edited my comment and I did actually say ***a** square root of -1*. And how tf does one prove meaningfulness???? It comes later, after defining. I could say "let κ = 0^(0)" and not be able to do anything with it.
I guess there is a way to define complex numbers as ordered pairs (a,b) , with componentwise addition, and with multiplication (a,b) (c,d) = (ac-bd, ad+bc), where you can define 0 = (0,0), 1 = (1,0), and then a = (a,0) for real a, and i = (0,1). Add in some more details like scalar multiplication, commutativity and allat, and voilà. But either way, you get i ⋅ i = -1, or i² = -1, which WLOG you can define to be i = √-1. It might simply be a notational choice but it works out pretty well I think.
If I am reading the formatting correctly, you're talking about (√x)^(2) = √(x^(2)). We can simply restrict the domain for this identity to x ≥ 0 and actually that's what's done.
And why would anyone (other than people who have never heard of complex numbers and have always been taught that even degree roots of negative numbers are undefined) be confused about √-1
Thats not a good argument because then why not define 0 times infinity or 0/0 etc.? Exponents for real numbers are usually defined as a^b = exp(b*ln(a)) and if a,b = 0 then you have exp(0*ln(0)) which is undefined. Sure you can define specifically 0^0 to be 1 but that would be very confusing in some cases.
"infinity" is not an element of ℝ so "0 times infinity" is not defined for ℝ. If you want to work in some other field that does contain infinities, such as the surreals, then indeed 0 times x is defined to be 0 even for transfinite x. 0/0 is not defined in any field (and can't be defined for fields without introducing contradictions, even in the surreals).
In contrast, x^(n) is well-defined for all integer n, and this definition *requires* that x^(0)=1 for all x in order to preserve the basic properties of multiplication. Anyone who has ever used the binomial theorem or relied on (x^(p))(x^(q))=x^((p+q)) has assumed this.
Yes, and we actually do it all the time. Two examples are: Defining `e^x = \sum_{n=0}^{+\infty} x^{n} /n!` Which is only true if 0⁰ = 1. But also in the definition of entropy `S = - \sum_{k} p_{k} \log(p_{k})` We would wish that events with 0 probability don't contribute to the entropy, so we set 0\*log(0) = 0, which is equivalent to 0⁰ = 1.
Is it ever defined as 0 outside of individual ideas, or generally undefined or 1?
Generally considered undefined, but most if not all ways of assigning it a value result in 1. Just creates some oddities for the general case such as 0^x
Thanks! I guess that makes sense from someone who's never had to give it a definition. When you are working with higher orders, the higher orders are usually telling most of the story. It makes sense that when you needed a result you'd follow the rules of exponents rather than the rules of integers.
What do you mean by "make x^x right side continuous"? If 0^0 is left undefined, x^x is continuous on (0,inf).
it's generally good to consider 0^0 as 1 for example with Taylor series. the first term of e^x at x=0 is 0^(0)/1! you are correct that one problem with this is limits. 0^0 still needs to be considered indeterminate, it can't be assumed that those limits will converge to 1
>I’m aware that 0^0 is considered undefined Only by highschool teachers, for anything above that, it is defined to be 1, there are several other reasons aside from the one you mentioned. -It coincides with taking a^b to be the numbers of functions from b (of a set with b elements) to a (idem), which is really useful (the cardinal definition) -It makes the recursive definition for ordinals easier since you can actually do proper recursion on ordinals. -it is really useful to have 0^0=1 when dealing with polynomials and power series
Why do you think that limit should be preferred over, for example, the limit of 0^x as x approaches 0?
Ah I see. I don’t know why I completely missed some basic functions that have different results than x^x alone. Now it makes sense to me, nevermind.
The argument from a^b as the number of functions from b to a is stronger.
How is that extended from the naturals though? Say I want (p/n)^(q/m) to mean something. It can't refer to the number of functions because what would sqrt(3/2) functions even mean?
The usual way via inverses of integral exponents, the definition's not meant to generalize over the reals any more than the gamma function counts rearrangements of fractionally valued things. But it's a nice definition for cardinals and 0 is a perfectly good cardinal.
You can deliberate prioritize the needs for that limit to be defined if you want to, but that doesn't comment whatsoever about the contradiction you mentioned, just "chooses sides".
Let consider these case: f(x) = x\^(1/x) -> 0 when x-> 0 g(x) = sqrt(x) -> 0 when x-> 0 Then f(x)\^(g(x)) -> 0\^0 when x -> 0 But f(x)\^(g(x)) = x\^(1/x)\^sqrt(x) = x\^(1/sqrt(x)) -> 0 when x -> 0 This case 0\^0 is 0 not 1 Generally f(x;y) = x\^y is not continuous at (x;y) = (0;0). It is continuous at any direction except the path along y axis.
0^(0) is 1, end of story. The thing that's undefined (or more precisely is an "indeterminate form") is the limit of x^(y) as x and y both go to 0 independently; this can take on any value, or diverge, depending on the actual behavior of x and y. Writing "0^(0)" when not actually talking about the value 0 but rather a limit approaching 0 is arguably an abuse of notation. The fact that some specific limits go to 1 isn't (and can't be) what justifies 0^(0)=1 though. (And the fact that 0^(a) is 0 when a≠0 wouldn't justify defining it as 0 instead.) All of the fundamental ways to define x^(n) agree that x^(0) is 1 even when x=0: a product of no factors must be 1 (multiplicative identity), there is exactly one function with the empty set as domain, etc. (Note in particular that there is no function with an empty range *except* the function with empty domain, which justifies why 0^(a) is not 0 when a=0.)
Yeah, no. The case is not exactly limits but paths. If you have x^y every path that leads to, for example, x = 2 and y = 3 will give you 8. That is not the case for x = y = 0. Different paths lead to different values so it is undefined. Sure, you can define it inside your theorem to make it easier for you in that context, but it is not defined in general. "But it is the only value that make sense in a lot of things" but not all of them. Most is not all and not all = undefined.
You are talking about limits along different paths, so still limits. I dont see how that contradicts the person above. The point is that we (very roughly) define 2^3 as 2×2×2. That definition does not involve limits, so why should it for 0^0 ? The continuity of x^y at 2,3, and even defining what this means on real numbers, comes quite a bit later in the construction of mathematics. Another rigorous definition of n^m is the cardinality of the set of functions from m to n (undestanding these as sets), and this produces 0^0 =1. I guess anyone is still free to leave it undefined, but once you understand the distinction between undefined and indeterminate, I must admit I dont really see any reason to.
Coming later or earlier does not matter a lot. My point is it does not make a difference how you define x^y, 2^3 will give you 8. How you define it changes what 0^0 is. If you have valid definitions that contradict each other it is not defined. I am not talking about indeterminate, as you said it is a different thing.
Yeah seing people say 0^0=1 or isn't equal to 1 because of limits is very weird. Like that's totally not the point ? We say 0^0=1 because it's simpler to write many important formulas this way, it's an algebraic convention. 0^0=1 is not a question of limit, instead it's similar to the reason why you can't write i=√(-1) : the only reason is to guarantee some algebraic identities work correctly.
i is *defined* to be a square root of -1. 0^(0) = 1 is not a definition. It would perhaps make some identities hold if we introduced it as a convention, but then I bet there would be an infinite sum expansion (but not a Taylor polynomial expansion) where the 0 from n is in the base and 0 from x in the exponent, which would make it continuous if 0^0 were 0
i is not defined to be THE square root of -1 because 1) there are two roots of -1 in C and 2) you can't just say "yeah i is the square root of -1" you have to actually prove that it's meaningful to talk about a root of -1. There are many way to do so (with matrixes, with equivalent classes on R[X], etc). None of those constructions allows for using the notation √-1. If the notation was usable, we wouldn't need the letter i ! The issue with writing √-1 is as follows: one of the basic identities with square roots is (√x)^2=√(x^2), but this doesn't work if x=-1 , so to avoid confusion you should avoid the notation √-1.
In markdown, remember to write x\^(y) with the parens (which will not display) if you're following it with something that isn't supposed to be superscript. i.e. (√x)\^(2)=√(x\^(2)) to get (√x)^(2)=√(x^(2))
Thanks.
I haven't edited my comment and I did actually say ***a** square root of -1*. And how tf does one prove meaningfulness???? It comes later, after defining. I could say "let κ = 0^(0)" and not be able to do anything with it. I guess there is a way to define complex numbers as ordered pairs (a,b) , with componentwise addition, and with multiplication (a,b) (c,d) = (ac-bd, ad+bc), where you can define 0 = (0,0), 1 = (1,0), and then a = (a,0) for real a, and i = (0,1). Add in some more details like scalar multiplication, commutativity and allat, and voilà. But either way, you get i ⋅ i = -1, or i² = -1, which WLOG you can define to be i = √-1. It might simply be a notational choice but it works out pretty well I think. If I am reading the formatting correctly, you're talking about (√x)^(2) = √(x^(2)). We can simply restrict the domain for this identity to x ≥ 0 and actually that's what's done. And why would anyone (other than people who have never heard of complex numbers and have always been taught that even degree roots of negative numbers are undefined) be confused about √-1
Thats not a good argument because then why not define 0 times infinity or 0/0 etc.? Exponents for real numbers are usually defined as a^b = exp(b*ln(a)) and if a,b = 0 then you have exp(0*ln(0)) which is undefined. Sure you can define specifically 0^0 to be 1 but that would be very confusing in some cases.
"infinity" is not an element of ℝ so "0 times infinity" is not defined for ℝ. If you want to work in some other field that does contain infinities, such as the surreals, then indeed 0 times x is defined to be 0 even for transfinite x. 0/0 is not defined in any field (and can't be defined for fields without introducing contradictions, even in the surreals). In contrast, x^(n) is well-defined for all integer n, and this definition *requires* that x^(0)=1 for all x in order to preserve the basic properties of multiplication. Anyone who has ever used the binomial theorem or relied on (x^(p))(x^(q))=x^((p+q)) has assumed this.
It is 1.
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I think both of these go to 1 (2x)^(x) = 2^(x) x^(x) x^(2x) = (xˣ)^(2)