7 ms·

A Quirky Function

83 points by mathgenius 10y ago

d4nte 10y ago
After reading the problem but before reading the solution, I came up with 0^ln(x+1). Did I do my math correctly, and is this a legitimate alternative answer?
- myle 10y ago
  x=0 has the same problem of 0^x being undefined
- jwmerrill 10y ago
  Yes, the results are the same for 0^f(x) where f(x) is any monotonic function with f(0)=0.
  - S4M 10y ago
    f has to be monotonic and increasing, and such as f(x) < 0 if x < 0 , and f(x) > 0 if x > 0 (note the strict inequalities, that makes it a condition slightly stronger than monotonic).
fisherjeff 10y ago
> And it equals its reciprocal, because 0^0 = 0^-0 = 1/0^0 Doesn't this already assume that 0^0 evaluates to 1 and not 0?
- qf303rjr3 10y ago
  No - you have the following identity because 0 = -0 0^0 = 0^(-0) and by the definition of negative exponents, 0^(-0) = 1/(0^0) So whatever number 0^0 is, it is equal to its own reciprocal.
  - fisherjeff 10y ago
    Ah, ok. I didn't have an issue with the first half, I just misunderstood: I thought he was stating that 0^0 == 1/0^0, not deriving it.
myle 10y ago
I think he is looking for Dirac Delta function.
- lvh 10y ago
  No, this is quite different (and the article tells you what it actually is). The Dirac delta is a distribution, this is a perfectly well-behaved function. It's just one that's tricky to write a formula for that doesn't require stepwise or partial functions.
cousin_it 10y ago
I like how deeply Baez thinks about the problem, but back when I was trying to figure it out, my reasoning was much simpler. Basically I just decided that keeping the binomial theorem true without exceptions was very important, so 0⁰ should be 1 by fiat. (If you set it to any other value x ≠ 1, then (0+1)¹ ≠ 0⁰⋅1¹ + 0¹⋅1⁰, because the left side is 1 and the right side is x.)
- GFK_of_xmaspast 10y ago
  It's Baez' blog, but the post was by David Tanzer: http://www.azimuthproject.org/azimuth/show/David+Tanzer http://www.azimuthproject.org/azimuth/show/David+Tanzer
tomp 10y ago
I studied math (BSc only, not PhD), and IMO that's not quite how math works. 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)).
- GFK_of_xmaspast 10y ago
  What's wrong with defining exp(z) in terms of the usual taylor series?
  - tomp 10y ago
    Yeah I guess that works as well. 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.
- JadeNB 10y ago
  Of course there is no unambiguously 'right' choice between two or more definitions that give the same answer, but I think that it is more common to switch your (1) and (2), and define ln(x) as \int_1^x dx/x.
S4M 10y ago
I got another one (a bit more complicated): find a function f that verifies: - 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)
- johnp_ 10y ago
  Not a math student, but I tried it anyway: 0 for x<=0 1 for x>=1 e^(1 - 1/(1 - (x - 1)^2)) else Does this work? Does the last condition mean smoothness (C∞)? (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))}]
  - S4M 10y ago
    That one will not work because the function is not C∞ [I didn't know how to make that symbol] in 1. it's first derivative is 0 in 1, but that's only because the derivative for 1-1/(1-(x-1)^2) is 0 in 1, that will not be the case for the second derivative. 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.
- kmill 10y ago
  https://en.wikipedia.org/wiki/Bump_function https://en.wikipedia.org/wiki/Bump_function
- GFK_of_xmaspast 10y ago
  My real analysis is extremely rusty, but could you use Urysohn's lemma to get an arbitrary continuous "step" function and then integrate that.
amelius 10y ago
> You can’t use any special functions, including things like sign and step functions, which are by definition discontinuous. 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?
- gizmo686 10y ago
  We can already define abs(x) as sqrt(x^2) = (x^2)^(1/2). 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 By composing this basic step function, we can construct an arbitrary step function. We can also note, that sign(x) comes very close to this definition. The only problem is that sign(0)=0. To work around this, we need a function, g, such that: g(x) = 1 | x=0 g(x) = 0 | x/=0 0^x comes close, but is not defined for x<0. To work around this limitation, we can take g(x)=0^(x^2). 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.
- jerf 10y ago
  "Is it possible to define a step-function in terms of this result?" 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.
dahart 10y ago
http://www.askamathematician.com/2010/12/q-what-does-00-zero-raised-to-the-zeroth-power-equal-why-do-mathematicians-and-high-school-teachers-disagree/ http://www.askamathematician.com/2010/12/q-what-does-00-zero... 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. ;)
- etatoby 10y ago
  > how about we try to make the function more continuous 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.
  - dahart 10y ago
    You're absolutely right. And you're also wrong in the very the same way I'm wrong. ;) We decided that 0^0=1, and by 'we', I mean the mathematicians. It's easy to show why the choice 0^0=1 is a bad choice, because it breaks some rules. It's also easy to show (as the article's author did) why it's a good choice too. But, regardless, it's still the choice that has been made, and the choice can't be unmade. It's a beautiful example, btw, thank you for contributing the visualization!
etatoby 10y ago
I've always marveled at the exponential function. 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 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 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.
- mathgenius 10y ago
  Also check out Greg Egan's visualization here: https://johncarlosbaez.wordpress.com/2016/06/25/a-quirky-function/#comment-80933 https://johncarlosbaez.wordpress.com/2016/06/25/a-quirky-fun...
- dahart 10y ago
  This is a great example. Wikipedia has the same one with isolines demonstrating this point in the 0^0 subsection of the exponentiation article, here: https://en.m.wikipedia.org/wiki/Exponentiation#Continuous_exponents 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." http://arxiv.org/pdf/math/9205211v1.pdf http://arxiv.org/pdf/math/9205211v1.pdf
nemo1618 10y ago
How about 1/(x * ∞)? Only works if you allow (0 * ∞) = 1, which is admittedly quite shaky.
- majewsky 10y ago
  All arithmetic with +/- ∞ is shaky. But even then, your solution is wrong because it's defined on negative numbers.
mjw_byrne 10y ago
This is an interesting discussion about why it isn't possible to define a "good" value for 0^0, i.e. a value which works in all cases, but the original riddle doesn't really work. "Nice" operations like addition, multiplication and exponentiation all preserve continuity and the function we're given is discontinuous. The given solution contains a hidden special case, rolled into the author's proposed definition for x^0. 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.
Double_Cast 10y ago
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)) Lambda Calculus returns (\z.z) AKA the Identity Function. Which kinda sounds like 1, except 1 is usually represented as (\sz.sz). Which means 0^0 is NaN? I wonder if division by zero also gives (\z.z), but I don't know what division in Lambda Calculus looks like.
- kmill 10y ago
  I'm pretty sure (\sz.sz) and (\z.z) are the same function: (\sz.(\z.z)sz)=(\sz.sz). 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).
kens 10y ago
The last time 0^0 came up on HN, I gave it a lot of thought. My conclusion is the problem goes away if you think in terms of typed functions. For the function x^y, where y is an integer, 0^0 is obviously 1. For the entirely different function x^y, where y is real (or complex), 0^0 is obviously undefined. The problem is people talk about 0^0 without making it clear which overloaded exponential function they are talking about. 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.
- Double_Cast 10y ago
  [Z] is a subset of [Q], which is a subset of [C]. So I don't think the problem goes away. 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.
  - kmill 10y ago
    For your first sentence, it really depends. If you start with the naturals, you can close it under addition using a Grothendieck-style construction: take pairs (n,m) of naturals, and say (n',m') is equivalent to it if n+m'=n'+m. There is a copy of the naturals embedded in these integers through n identified as (n,0). 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.
joefkelley 10y ago
Here's Knuth's opinion on 0^0: http://arxiv.org/pdf/math/9205211v1.pdf http://arxiv.org/pdf/math/9205211v1.pdf (on page 6) 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.
dahart 10y ago
Check it out - this has been debated for at least 200 years! https://en.m.wikipedia.org/wiki/Exponentiation#History_of_differing_points_of_view https://en.m.wikipedia.org/wiki/Exponentiation#History_of_di...

83 points by mathgenius 10y ago

d4nte 10y ago

After reading the problem but before reading the solution, I came up with 0^ln(x+1). Did I do my math correctly, and is this a legitimate alternative answer?

myle 10y ago
x=0 has the same problem of 0^x being undefined
jwmerrill 10y ago
Yes, the results are the same for 0^f(x) where f(x) is any monotonic function with f(0)=0.
- S4M 10y ago
  f has to be monotonic and increasing, and such as f(x) < 0 if x < 0 , and f(x) > 0 if x > 0 (note the strict inequalities, that makes it a condition slightly stronger than monotonic).

fisherjeff 10y ago

> And it equals its reciprocal, because 0^0 = 0^-0 = 1/0^0 Doesn't this already assume that 0^0 evaluates to 1 and not 0?

qf303rjr3 10y ago
No - you have the following identity because 0 = -0 0^0 = 0^(-0) and by the definition of negative exponents, 0^(-0) = 1/(0^0) So whatever number 0^0 is, it is equal to its own reciprocal.
- fisherjeff 10y ago
  Ah, ok. I didn't have an issue with the first half, I just misunderstood: I thought he was stating that 0^0 == 1/0^0, not deriving it.

myle 10y ago

I think he is looking for Dirac Delta function.

lvh 10y ago
No, this is quite different (and the article tells you what it actually is). The Dirac delta is a distribution, this is a perfectly well-behaved function. It's just one that's tricky to write a formula for that doesn't require stepwise or partial functions.

cousin_it 10y ago

I like how deeply Baez thinks about the problem, but back when I was trying to figure it out, my reasoning was much simpler. Basically I just decided that keeping the binomial theorem true without exceptions was very important, so 0⁰ should be 1 by fiat. (If you set it to any other value x ≠ 1, then (0+1)¹ ≠ 0⁰⋅1¹ + 0¹⋅1⁰, because the left side is 1 and the right side is x.)

GFK_of_xmaspast 10y ago
It's Baez' blog, but the post was by David Tanzer: http://www.azimuthproject.org/azimuth/show/David+Tanzer http://www.azimuthproject.org/azimuth/show/David+Tanzer

tomp 10y ago

I studied math (BSc only, not PhD), and IMO that's not quite how math works. 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)).

GFK_of_xmaspast 10y ago
What's wrong with defining exp(z) in terms of the usual taylor series?
- tomp 10y ago
  Yeah I guess that works as well. 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.
JadeNB 10y ago
Of course there is no unambiguously 'right' choice between two or more definitions that give the same answer, but I think that it is more common to switch your (1) and (2), and define ln(x) as \int_1^x dx/x.

S4M 10y ago

I got another one (a bit more complicated): find a function f that verifies: - 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)

johnp_ 10y ago
Not a math student, but I tried it anyway: 0 for x<=0 1 for x>=1 e^(1 - 1/(1 - (x - 1)^2)) else Does this work? Does the last condition mean smoothness (C∞)? (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))}]
- S4M 10y ago
  That one will not work because the function is not C∞ [I didn't know how to make that symbol] in 1. it's first derivative is 0 in 1, but that's only because the derivative for 1-1/(1-(x-1)^2) is 0 in 1, that will not be the case for the second derivative. 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.
kmill 10y ago
https://en.wikipedia.org/wiki/Bump_function https://en.wikipedia.org/wiki/Bump_function
GFK_of_xmaspast 10y ago
My real analysis is extremely rusty, but could you use Urysohn's lemma to get an arbitrary continuous "step" function and then integrate that.

amelius 10y ago

> You can’t use any special functions, including things like sign and step functions, which are by definition discontinuous. 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?

gizmo686 10y ago
We can already define abs(x) as sqrt(x^2) = (x^2)^(1/2). 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 By composing this basic step function, we can construct an arbitrary step function. We can also note, that sign(x) comes very close to this definition. The only problem is that sign(0)=0. To work around this, we need a function, g, such that: g(x) = 1 | x=0 g(x) = 0 | x/=0 0^x comes close, but is not defined for x<0. To work around this limitation, we can take g(x)=0^(x^2). 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.
jerf 10y ago
"Is it possible to define a step-function in terms of this result?" 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.

dahart 10y ago

http://www.askamathematician.com/2010/12/q-what-does-00-zero-raised-to-the-zeroth-power-equal-why-do-mathematicians-and-high-school-teachers-disagree/ http://www.askamathematician.com/2010/12/q-what-does-00-zero... 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. ;)

etatoby 10y ago
> how about we try to make the function more continuous 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.
- dahart 10y ago
  You're absolutely right. And you're also wrong in the very the same way I'm wrong. ;) We decided that 0^0=1, and by 'we', I mean the mathematicians. It's easy to show why the choice 0^0=1 is a bad choice, because it breaks some rules. It's also easy to show (as the article's author did) why it's a good choice too. But, regardless, it's still the choice that has been made, and the choice can't be unmade. It's a beautiful example, btw, thank you for contributing the visualization!

etatoby 10y ago

I've always marveled at the exponential function. 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 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 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.

mathgenius 10y ago
Also check out Greg Egan's visualization here: https://johncarlosbaez.wordpress.com/2016/06/25/a-quirky-function/#comment-80933 https://johncarlosbaez.wordpress.com/2016/06/25/a-quirky-fun...
dahart 10y ago
This is a great example. Wikipedia has the same one with isolines demonstrating this point in the 0^0 subsection of the exponentiation article, here: https://en.m.wikipedia.org/wiki/Exponentiation#Continuous_exponents 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." http://arxiv.org/pdf/math/9205211v1.pdf http://arxiv.org/pdf/math/9205211v1.pdf

nemo1618 10y ago

How about 1/(x * ∞)? Only works if you allow (0 * ∞) = 1, which is admittedly quite shaky.

majewsky 10y ago
All arithmetic with +/- ∞ is shaky. But even then, your solution is wrong because it's defined on negative numbers.

mjw_byrne 10y ago

This is an interesting discussion about why it isn't possible to define a "good" value for 0^0, i.e. a value which works in all cases, but the original riddle doesn't really work. "Nice" operations like addition, multiplication and exponentiation all preserve continuity and the function we're given is discontinuous. The given solution contains a hidden special case, rolled into the author's proposed definition for x^0. 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.

Double_Cast 10y ago

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)) Lambda Calculus returns (\z.z) AKA the Identity Function. Which kinda sounds like 1, except 1 is usually represented as (\sz.sz). Which means 0^0 is NaN? I wonder if division by zero also gives (\z.z), but I don't know what division in Lambda Calculus looks like.

kmill 10y ago
I'm pretty sure (\sz.sz) and (\z.z) are the same function: (\sz.(\z.z)sz)=(\sz.sz). 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).

kens 10y ago

The last time 0^0 came up on HN, I gave it a lot of thought. My conclusion is the problem goes away if you think in terms of typed functions. For the function x^y, where y is an integer, 0^0 is obviously 1. For the entirely different function x^y, where y is real (or complex), 0^0 is obviously undefined. The problem is people talk about 0^0 without making it clear which overloaded exponential function they are talking about. 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.

Double_Cast 10y ago
[Z] is a subset of [Q], which is a subset of [C]. So I don't think the problem goes away. 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.
- kmill 10y ago
  For your first sentence, it really depends. If you start with the naturals, you can close it under addition using a Grothendieck-style construction: take pairs (n,m) of naturals, and say (n',m') is equivalent to it if n+m'=n'+m. There is a copy of the naturals embedded in these integers through n identified as (n,0). 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.

joefkelley 10y ago

Here's Knuth's opinion on 0^0: http://arxiv.org/pdf/math/9205211v1.pdf http://arxiv.org/pdf/math/9205211v1.pdf (on page 6) 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.

dahart 10y ago

Check it out - this has been debated for at least 200 years! https://en.m.wikipedia.org/wiki/Exponentiation#History_of_differing_points_of_view https://en.m.wikipedia.org/wiki/Exponentiation#History_of_di...