A Quirky Function
johncarlosbaez.wordpress.com
johncarlosbaez.wordpress.com
Here's a 3D visualization of z=x^y that supports the 0^0 is undefined camp:
https://s32.postimg.org/v9t4stdl1/x_y.gif
(Made with Grapher.app and Gimp)
You can clearly see that the surface is tangent to the z (vertical) axis, which suggests that any z>=0 satisfies z=0^0, hence 0^0 is undefined.
Edit: here's a picture that highlights two important rules:
https://s31.postimg.org/4kozbjqbf/x_y_lines.png
The red line is the rule 0^y=0 for all y; the purple line is x^0=1 for all x. It's clear why these two lines would not meet at 0^0. Some will define 0^0=0, others 0^0=1. But the 3D surface shows that it's really a vertical line.
https://en.m.wikipedia.org/wiki/Exponentiation#Continuous_ex...
With the caveat:
"However, under certain conditions, such as when f and g are both analytic functions and f is positive on the open interval (0, b) for some positive b, the limit approaching from the right is always 1.[26][27][28]"
Also, definitely take note of Knuth's argument, because while you might feel strongly that indeterminate is the right answer right now, there are big implications for allowing other choices. (And NB: indeterminate vs undefined)
"The debate stopped there, apparently with the conclusion that 0^0 should be undefined.
But no, no, ten thousand times no! Anybody who wants the binomial theorem to hold for at least one nonnegative integer n must believe that 0^0 = 1, for we can plug in x = 0 and y=1 to get 1 on the left and 0^0 on the right. The number of mappings from the empty set to the empty set is 0^0. It has to be 1."
https://johncarlosbaez.wordpress.com/2016/06/25/a-quirky-fun...
It might be a mistake to try to argue that defining 0^0=1 is the best answer or the most consistent answer. The problem is that there are conflicts. There is no such thing as more consistent, if any rule breaks, they all break. Some math rules lead you to 0^0=0, some to 0^0=1, and some to 0^0=undefined. So math doesn't have an answer for this, and what really happens is we decide that 0^0=1 by convention. There are other examples of this happening in math, and it works reasonably well. But let's not pretend it's "right", let's accept that it's a choice.
We're choosing to say that the rule x^0=1 takes precedence over the rule 0 * x = 0.
Just to play devil's advocate, I like the choice of 0 for intuition consistency, not math consistency. The positive side limit is 0, so if 0^x is defined at x=0, how about we try to make the function more continuous, instead of trying to make it discontinuous?
0^0 is also saying 'start with zero, then multiply it by itself ... Wait, on second thought, don't start'
There's nothing, because you don't multiply zero by anything. The best number we have to represent nothing is zero. Saying that not raising zero to a power is one feels like a random number.
But, I'm wrong. ;)
The same is true for any other integer: "2^0 is also saying 'start with two, then multiply it by itself ... Wait, on second thought, don't start'" You don't multiply two by anything.
http://math.stackexchange.com/questions/9703/how-do-i-explai...
The thing is, there is no argument to be had, x^0 is defined to be 1, and x^1 is defined to be x. Those agree with and make consistent lots of other rules and limits, but they are choices we made just like saying 0^0=1.
x^0 = x^(y - y) = x^y * x^-y = x^y * 1/x^y = x^y / x^y = 1
x^1 isn't defined to be x, x^1 IS x.
The situation is analoguous to multiplication. a X b is defined as a+a+a+... b times. If you see a problem with x^0 and x^1, you should also see a problem with a X 0 and a X 1.
x^0 is 1 because 1 is multiplication's identity element, just like a X 0 is zero because zero is addition identity element: https://en.wikipedia.org/wiki/Identity_element
So you say. If it's planly obvious, why do I get so many results for the Google search "why is x^0 1"?
I understand the multiplicative identity argument, but that's a technical explanation, not an intuitive one.
What's the intuitive reason that multiplying something zero times should equal one? If I don't multiply, then I don't have an answer, and zero's closer to nothing than one is. Why should I use the multiplicative identity if I don't multiply, why would that make sense?
Using the multiplicative identity is yet another choice, not an intrinsic property of numbers. It's a very good choice, and there are a lot of reasons why it's a good choice.
https://en.m.wikipedia.org/wiki/Empty_product
"In mathematics, an empty product, or nullary product, is the result of multiplying no factors. It is by convention equal to the multiplicative identity 1 (assuming there is an identity for the multiplication operation in question), just as the empty sum—the result of adding no numbers—is by convention zero, or the additive identity.[1][2][3]"
> x^0 isn't defined to be 1, x^0 IS 1.
Are you sure? How can you show that? How do you define what an exponent is without defining what x^0 and x^1 are? What does it mean to suggest that x^0 is intrinsically 1? Are you absolutely certain that you just aren't so comfortable with the idea that you can't imagine other possibilities?
https://en.m.wikipedia.org/wiki/Exponentiation
"Formally, powers with positive integer exponents may be defined by the initial condition
b^1 = b
and the recurrence relation
b^(n+1) = b^n * b"
Now, since I just quoted the definition of exponentiation itself from Wikipedia and the formal definition includes the base case b^1 = b, I think you've completely failed to make an educated point to go along with your insult. I didn't ask for you to define exponentiation so you could educate me, I asked so that you could think carefully and tell me if you can define what exponentiation is without using the base case. Were you to actually try, you may find it difficult. Or not, I might be wrong, so feel free to prove me wrong or cite a source that proves me wrong, if you want. As it stands, my takeaway for now is that your insult is a substitute for the argument you don't have, so you're forfeiting your position and handing me a walkover.
x^y / x^y
If you wanted to prove that 0^0 really does equal 1, you would have to prove that the output of the reduction is unique.
Look at the 3D rendering in my comment. When you look at the z=x^y surface and try to make it continuous, what comes out at 0^0 is a vertical line, aka. undefined.
It's a beautiful example, btw, thank you for contributing the visualization!
https://s31.postimg.org/4kozbjqbf/x_y_lines.png
that highlights 0^y=0 in red and x^0=1 in purple. Now it's even more obvious why any single answer is insufficient: 0^0 is a vertical line, not a single point!
Your re-iterating your point makes me wonder if you're actually hearing and understanding the part about choices and definitions trumping logic. If we define 0^0=1, then it doesn't matter what arguments we have, it doesn't matter that there are 3 or more right answers. The right answer can be, and in other parts of math, is what we define it to be, not what makes the most sense.
0^0 is defined as 1 (sometimes) because that has utility and consistency for some other expressions, formulas, operations, etc. It's not because it's right. It's not because it makes sense. It's because we decide to set 0^0=1. It's for convenience.
Now, some sources online say that 0^0 is indeterminate, not equal to one. Other sources online say that 0^0 is 1.
http://mathworld.wolfram.com/Power.html https://en.wikipedia.org/wiki/Indeterminate_form
https://cs.uwaterloo.ca/~alopez-o/math-faq/mathtext/node14.h... https://www.google.com/search?q=0%5E0 https://betterexplained.com/articles/understanding-exponents...
From a formal mathematical standpoint, you define things in certain ways and then work with the definitions. It's not like there is a "correct" value for 0^0 out there that we can mount a search for.
It's roughly the same with discussion of 0.999... - people still debate whether it's less than or equal to 1. You just have to use the definitions. In this case, 0.999... is shorthand for an infinite sum. An infinite sum is the limit of the sequence of its partial sums (where such exists). In this case the limit is provably one. So that's the value of 0.999... . If you look at a series like 1 - 1 + 1 - 1 + ..., its sequence of partial sums is divergent, so it doesn't have a value and that's that - although that doesn't mean we can't apply all sorts of alternative definitions to try to work with such a series, and some of them do in fact assign it a value. Such alternative definitions can produce really weird-looking results - one of the most striking is the result that (subject to certain definitions) 1 + 2 + 3 + 4 + ... is -1/12.
If you're doing things with x^n (integer n), such as combinatorics or polynomials, you need to have 0^0=1 or else you have lots of annoying special cases. e.g. if you have a polynomial sum(an * x^n), you don't want to have it break at x=0.
But if you're working with real or complex exponents, defining 0^0=1 is just strange.
If you examine the graph etatoby provided, the limit of 0^0 depends on which direction it's approached from: 1 from along the real axis; 0 from along the imaginary axis; and all sorts of [Q] from along a diagonal.
Similarly, we can take pairs of these integers to close under division to get the rationals. Let's denote such pairs n/m (m nonzero), so n'/m' is equivalent when nm'=n'm. The integers are represented as a pair by n/1.
Next we can close the rationals under the property that Cauchy sequences converge: take the set of all Cauchy sequences of rationals, and say two such sequences are equivalent if the term-wise difference between the two converges to the rational 0/1. This gives us the reals, with a rational q being identified with the sequence (q,q,q,...)
Finally, we can close the reals under being able to have polynomials have roots. It turns out adding the root of x^2+1 is all you need. The complex numbers then are the set of all polynomials with real coefficients, where two polynomials are equivalent if their difference is divisible by x^2+1. The reals can be identified as a complex number by thinking of r as a constant polynomial.
All I'm saying is that, while we commonly think of Z as a subset of Q as a subset of R as a subset of C, this is only after a sequence of closures of different kinds, each time creating a completely new set which the first number system is not actually a subset of (though the first number system can be identified as some subset in the new system).
With this in mind, it's completely reasonable to define exponentiation of something to the integer or to the rational as being a completely different operation from exponentiation to the real or to the complex. In fact, in Rudin, exponentiation to the real is defined as a limit of exponentiation to the rationals, which is defined as a positive root of real to the integer.
Why is the integer case obvious? I feel like the absolute simplest example juxtaposing 1^0=1 with 0^1=0 is enough to make figuring 0^0 obviously ambiguous. :)
Considering the integer exponent and real exponent cases separately isn't just a crazy idea of mine, the IEEE floating point spec does it too. The pown function is ^ where the second value is an integer and powr is ^ where the second value is real. (I'm ignoring the difference between doubles and reals.)
Why should pown(0, 0) be 1? pown(x, 0) is 1 for all x other than 0. If pown(0,0) is not 1, it breaks the binomial formula and lots of other convenient things. There's no other relevant limit: pown(0, -1) is undefined and pown(0, 1) is 0, and there's nothing in between. So if you're thinking of ^ as pown (e.g. you're a discrete mathematics person), 0^0 is 1.
On the other hand powr(0, 0) should be undefined. As etatoby's graph shows, the limit depends how you approach it. So if you're thinking of ^ as powr (e.g. you're a real analysis person), 0^0 is undefined.
To conclude, the IEEE floating point spec uses 2 (actually 3 for historical reasons) different functions for x^y. The value of 0^0 depends on which function you use. Arguing over the value of 0^0 is basically arguing whether pown or powr is the "right" function for x^y, and it obviously depends on the situation.
(To clarify: I'm not arguing that it should be done this way because the IEEE spec does it; I'm using the IEEE spec to try to explain the two cases.)
I'm not sure I understand what you mean there.
For any positive nonzero real value x, x^0=1, right? And 0^x=0 for all x too, yes?
The limit from the positive direction for x^0 is 1, you can get arbitrarily close to zero. The limit from the positive direction for 0^x is 0, and you can also get arbitrarily close to 0 here too.
etatoby's graph with isolines, the red and blue ones, shows those two cases specifically, and yeah, they arrive at 0 with two different values, meaning the answer is indeterminate (but not undefined).
Etatoby is right; 0^0 is indeterminate. I'm not arguing with that at all, it is a fact.
I'm saying I understand why people choose to ignore that 0^0 is indeterminate and choose instead to just define that 0^0=1.
I really like your pown() vs powr() example, because it actually demonstrates how in practice, choosing to return a defined value in some cases is the more practical choice.
That said, I think there's still a conflict even within pown(), even for the integer case, that's what I was getting at. pown(0,x) is 0 for all positive x except 0. Negative x is undefined due to the division by 0, and pown(0,0)=1 simply because 0^0 is chosen to be defined at 1. But if you imagine an integer limit starting from 10 going down to zero, there's a clear pattern:
pown(0,10)=0; pown(0,9)=0; ... pown(0,1)=0
pown(10,0)=1; pown(9,0)=1; ... pown(1,0)=1
IEEE chose to make the pown(x,0) rule more important than the pown(0,x) rule when it comes to x=0, even in the case of integers.
It feels wrong to introduce a distinction.
You can define a^x or 0^x by its algebraic properties only for rational numbers (i.e. fractions). In algebra, it makes sense that 0^0 = 1, but that won't allow you to plot 0^x as a function.
To plot it as a function, you need to define it for all real numbers, which, AFAIK, you need to do analytically:
1. First, define e^x as the limit (1 + x/n)^n as n goes to infinity (n is a natural number, so this definition uses the standard algebraic exponentiation).
2. Then, you define ln(x) as the reverse of e^x (after proving that e^x is monotonous and continuous).
3. Then, define a^x as e^(x * ln a).
4. Then, 0^x is defined as a^x as a -> 0, which is 0 for x > 0. In the same way, you can see that x^0 is 1 for x > 0. Therefore, there is no "correct" (analytically) result for 0^0 (i.e. such that would make the function f(x, y) = x ^ y continuous at (x, y) = (0, 0)).
I wonder though how hard it would be to prove basic equalities (e.g. e^(x + y) = e^x * e^y)... Probably not too hard since that's how you define the complex exponential function.
0^0
(\be.eb)(\sz.z)(\sz.z)
[b := (\sz.z)]
(\e.e(\sz.z))(\sz.z)
[e := (\sz.z)]
((\sz.z)(\sz.z))
[s := (\sz.z)]
((\z.z))
It would make sense that 0^0 for church numerals is 1, since this is exponentiation to the natural (defined as recursive multiplication, with the base case x^0=1).
So, let's go the other way around. Is it possible to define a step-function in terms of this result? Can we define functions such as sign(x) and abs(x) in terms of it?
sign(x) follows naturally as abs(x)/x.
The step function is a bit trickier, we can solve this with 0^0=1.
Begin by noting, that to construct an arbitrary step function, all we need is a function f such that:
f(x) = -1 | x<0
f(x) = 1 | x>= 0
To work around this, we need a function, g, such that:
g(x) = 1 | x=0
g(x) = 0 | x/=0
With this, we find that f(x) = sign(x) + 0^(x^2), satisfies our initial criteria. By composing functions of the form `a * f(x + b)`, we can construct an arbitrary step function.
It might be possible to define it in terms of the semi-arbitrary choice of what 0^0 is, but that's generally mathematically unsatisfactory, almost certainly leading to no non-equally-arbitrary insights.
He makes a few pretty convincing points about why it's useful to define it equal to 1, while admitting that in some sense, it's not quite as well defined as say 0+0.
Doesn't this already assume that 0^0 evaluates to 1 and not 0?
0^0 = 0^(-0)
0^(-0) = 1/(0^0)
https://en.m.wikipedia.org/wiki/Exponentiation#History_of_di...
- f(x) = 0 for x <= 0
- f(x) = 1 for x >= 1
- f admits a derivative of any order
The last condition rules out functions like f(x) = 0 if x<=0,x if x in [0,1] and 1 if x >=0.
(Note: piecewise defined functions are still OK)
0 for x<=0
1 for x>=1
e^(1 - 1/(1 - (x - 1)^2)) else
(Not a native english or math speaker, so please forgive my ignorance)
edit: HN ate the link :/ wolframalpha code:
Piecewise[{{0, x <= 0}, {1, x >=1}}, {e^(1 - 1/(1 - (x - 1)^2))}]If you look at what's going on in 0, the derivative of any order is something like phi(x) e^(1 - 1/(1 - (x - 1)^2)) where phi is a rational function, and thus it will always be 0 in 0 (the exponential always beats a rational function).
so the answer is, similarly to what kmill posted:
0 for x <=0 1 for x >= 1 1/(1+exp(K(x))) else
With K(x) = (x-1/2)/(x(x-1))
(x-1/2 is added to create a change of sign between 0 and 1)
in 0, K(x) -> ∞, so 1/(1+exp(K(x)) -> 0, and in 1, K(x) -> -∞, so 1/(1+exp(K(x)) -> 1.
To see that the function is C∞, you can check that the n-th derivative 1/(1+exp(K(x))) is of the form phi(x) exp(K(x)) 1/(1+exp(K(x)))^(n+1) where phi is a rational function, and then verify that the n-th derivative is 0 in 1 and 0.
Only works if you allow (0 * ∞) = 1, which is admittedly quite shaky.