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March 30

0^0 (cont'd!)

I have read the previous discussions on the definition of $0^{0}$. The controversy arises solely because the limit $\lim _{(x,y)\rightarrow (0,0)}x^{y}$ does not exist... but does it matter that it doesn't exist? What's wrong with simply defining $0^{0}=1$ and acknowledging that the function $x^{y}$ is discontinuous at $(0,0)$? 101.119.129.156 (talk) 13:40, 30 March 2025 (UTC)

Maybe it is because mathematicians like to be purists. Although typical real-world problems consider $0^{0}=1$, there are theoretical systems where the limit exists but has a different value. 101.119.129.156 (talk) 14:00, 30 March 2025 (UTC)

Continuity matters because operations on real numbers require it. Consider the expression π√2. You can't say "Add π to itself √2 times" because that's nonsense. Instead you have to build up the definition from integers to rationals, then from rationals to real numbers. In detail, first define r⋅n as r added to itself n times; this can be done inductively: r⋅0 = 0, r⋅(n+1) = r⋅n + r. Then define r⋅(a/b) when a and b are integers, as the solution to p⋅b=r⋅a. Finally define x⋅y as the limit of (a_i/b_i)(c_i/d_i) where a_i/b_i has limit x and c_i/d_i has limit y. But without knowing that r⋅s is a continuous function of r and s you can't guarantee that your limiting value of (a_i/b_i)(c_i/d_i) doesn't depend on which sequences a_i/b_i and c_i/d_i you're using. Multiplication is continuous so there is no such problem extending the definition from rationals to reals. But exponentiation is not continuous so there is a problem. You have to restrict the domain of the operation so that this issue does not arise. And this has been done to extend the definition as much as possible, though this becomes awkward to state concisely. If you restrict r to positive values then r^s is continuous and can be extended to x^y for real x and y as long as x is positive. If n is a non-negative integer then rⁿ is continuous for all r, so xⁿ can be defined for all real x. Because r^s is not a continuous function in the neighborhood of r=0, s=0, the expression 0⁰ is problematic when considered as the case r=0, s=0 of the expression r^s. It's not that mathematicians like to be purists, but that they like to have expressions mean something definite and not be a matter of opinion. For more detail, the relevant article is Zero to the power of zero. --RDBury (talk) 20:22, 30 March 2025 (UTC)

Here is a concrete example. Take the problem of determining

$\lim _{x\downarrow 0}f(x)^{g(x)}$

and consider the rule

$\lim _{x\downarrow 0}f(x)^{g(x)}=F^{G},~{\text{where}}~F=\lim _{x\downarrow 0}f(x)~{\text{and}}~G=\lim _{x\downarrow 0}g(x).$

This seems reasonable enough. Now apply this to

$f(x)=\exp \left(-{\frac {1}{x}}\right),~g(x)=x.$

In this case $F=G=0,$ so the rule suggests that the answer is $0^{0}.$ But the actual limit is ${\frac {1}{e}}\approx 0.37\,.$

When the exponent is restricted to the domain of the natural numbers, the notion of it having a limit does not apply, so then there is no ground for considering $0^{0}$ an indeterminate form, and defining $x^{0}=1$ without restriction is perfectly reasonable. With that convention, defining a Taylor series by $\textstyle {\sum _{i=0}^{\infty }}a_{i}x^{i}$ means the same as, but is more convenient than, $a_{0}+\textstyle {\sum _{i=1}^{\infty }}a_{i}x^{i}$  ​‑‑Lambiam 21:04, 30 March 2025 (UTC)

---

To be honest, I rather agree with the OP (and Don Knuth, p. 6). There is nothing wrong with it, and it is a perfectly defensible convention. For me, just as you say, $0^{0}=1$ by definition (so that the binomial theorem works without special cases), and you just need to be aware that $f(x,y)=x^{y}$ is not continuous at (0,0). It is also just like any empty product: you are multiplying no numbers.

Also I am not aware of any common reason to define $0^{0}$ as anything specific other than 1, if you define it at all. Double sharp (talk) 12:52, 31 March 2025 (UTC)

I think RDBury's answer is a good one. Saying "$0^{0}=1$ in the integers, therefore it should have that definition in any context" is a bit like saying "${\frac {0}{0}}=0$ in the trivial group, therefore it should have that definition in any context". Unless I'm missing something. 101.119.121.180 (talk) 13:46, 31 March 2025 (UTC)

Well $\mathbf {N} \subset \mathbf {R} \subset \mathbf {C}$, whereas the trivial group doesn't have that going for it. And you're implicitly saying $0^{0}=1$ if you want to write polynomials or power series in the form $f(x)=\sum _{k}a_{k}x^{k}$ and write $f(0)=a_{0}$. Not to mention if you want the binomial theorem to hold without awkward edge cases, e.g. $0^{0}=(-1+1)^{0}={\binom {0}{0}}(-1)^{0}1^{0}=1$. Double sharp (talk) 13:57, 31 March 2025 (UTC)

Uh, I'm pretty sure the trivial group is a subset of $\mathbf {N}$. 101.119.121.180 (talk) 14:04, 31 March 2025 (UTC)

True. What I meant (but failed hilariously at saying) is that if you think of the trivial group as $\{0\}$ under multiplication, then the multiplications are incompatible (since you can't sensibly define $0^{-1}$ in $\mathbf {N}$). Whereas saying $0^{0}=1$ in $\mathbf {N}$ doesn't really break exponentiation in $\mathbf {R}$: sure it becomes discontinuous, but anything you do at $0^{0}$ will make it so, so why not just use the convenient value? Double sharp (talk) 14:06, 31 March 2025 (UTC)

RDBury's answer is a good one like that person said. There is absolutely no good reason to extend the natural number/set theory version to real numbers, it just causes problems. In analysis it is much better to say something does not have a limit or to calculate a limit than to mandate some discontinuous value it's just asking for trouble. NadVolum (talk) 17:14, 31 March 2025 (UTC)

I guess this is where we are going to disagree, then. I think, following Knuth, that the binomial theorem and being able to write power series compactly are important enough for analysis that it's better to mandate the discontinuous value $0^{0}=1$: the alternative (having to put special cases in the binomial theorem) just seems even worse.

Anyway, if we can deal in analysis with step functions with their discontinuities, then what is wrong with having a discontinuous exponential function? It is not as if $0^{0}$ can be given any value that makes $f(x,y)=x^{y}$ continuous. We can, however, choose to give it the one value that makes sense in some context. Double sharp (talk) 18:04, 31 March 2025 (UTC)

The polynomial case has a natural-number exponent. It does make sense to define the function with a real base and a natural-number exponent in such a way that $0.0^{0}=1.0$. What does not have such a clear motivation is to define the function with real base and real exponent in the case of $0.0^{0.0}$. --Trovatore (talk) 18:12, 31 March 2025 (UTC)

Well, then let's just go with the binomial theorem. That certainly allows the exponent to be real. Or indeed, complex. Surely we would not claim that $(-1)^{0}$ should be left undefined just because we can't speak of continuity here unless we go straight to $\mathbf {C}$? Or should we start leaving $0^{1}$ undefined because the real exponential is conceptually $f(x,y)=\exp(y\ln x)$, and $\ln 0$ is undefined? Or are we going to start worrying about $0^{\pi }$, since there we don't even have the discrete exponential to fall back to? I dunno, I think it's most convenient just to expand the domain of the exponential as far as possible by saying that $0^{z}$ equals 0 when $\Re (z)>0$ (by limiting arguments), and that $0^{0}=1$ as a special case. I cannot think of a situation when these values are "wrong" in the sense that it gets harder to state theorems if you use them; you just need to be aware of the discontinuity. Double sharp (talk) 18:13, 31 March 2025 (UTC)

I am not sure what instance of the binomial theorem you have in mind. Yes, the domain of the exponential function as a partial function from $\mathbf {R} \times \mathbf {R} \to \mathbf {R}$, $(x,y)\mapsto x^{y}$, is $x>0$. There is really no particular use I am aware of to extend the function past that. --Trovatore (talk) 20:08, 31 March 2025 (UTC)

@Trovatore: Following Knuth, I mean this: $1=(0+1)^{r}=\sum _{k=0}^{\infty }{\binom {r}{k}}0^{k}1^{r-k}$. All the terms are clearly 0 except the first, which is $0^{0}$. It is for this kind of thing that I think it's worth defining $0^{0}=1$ explicitly: refusing to do so means you need to pedantically exclude special cases for the binomial theorem. As for $0^{\pi }$, maybe it is indeed not something you'd ever really need, but the answer $0$ is both obvious and doesn't break anything. So, really, why not? Double sharp (talk) 20:14, 31 March 2025 (UTC)

The first term there is $0.0^{0}$, not $0.0^{0.0}$. --Trovatore (talk) 22:59, 31 March 2025 (UTC)

Nothing is 'wrong' with that convention: you are free to write it down and use it all you like. You can even fire up Python, type print(0 ** 0), and get the answer 1; or type print(0.0 ** 0.0), and get the answer 1.0. But it does not make sense in all contexts. For that reason, the IEEE 754 rules allow three different kinds of exponentiation, which make different choices here: [1], [2]. --Amble (talk) 19:44, 31 March 2025 (UTC)

March 31

Source for calculation of singular vectors

I am looking for a step-by-step guide to calculating singular vectors. This is what I have so far: Given a matrix A, which is not square, calculate M as A matrix multiplied to A' (A transposed). It can be A'A or AA'. Does not matter for the final result. Calculate the eigenvectors of M as e1, e2, etc... ??? You have singular vectors. In attempting to fill in the ??? part, every web search shoves me to singular value decomposition. I am not attempting to calculate a singular value decomposition. I am trying to calculate the singular vectors of the original non-square matrix A.

For an example. I am attempting to calculate the singular vectors for the four data sets from Anscombe's quartet. For set 1, when I multiply A'A, I get the matrix [[1001, 797.6], [797.6, 660.17]]. I did the math and got the two eigenvalue,vector pairs: 1646.19 [1.24, 1] and 14.98 [-0.81,1]. I don't know what the singular vectors are. My understanding is the singular values are the square roots of the eigenvalues. So, the square root of 1646 is 40.57 and the square root of 14.98 is 3.87. But, I want the singular vectors, not the singular values. 68.187.174.155 (talk) 16:53, 31 March 2025 (UTC)