Zero raised to a positive real number

[psie]

TL;DR

I'm browsing through Rudin's Principles of Mathematical Analysis and I'm wondering if he ever defines or if it's otherwise possible to deduce from any of the theorems in the book what zero raised to a positive real number equals.

I feel silly for asking, since I have accepted this always as true, but I don't have a reference for what $0^p$ equals when $p$ is a positive real number. This dawned on me when trying to show the positive definiteness of the $p$-norm for $x\in\mathbb R^n$, that is, $$x=0\iff \left(\sum_1^n |x_j|^p\right)^{1/p}=0.$$ Here $1\leq p<\infty$. I simply realized I can't prove this equivalence unless I accept that $0^p=0$. I can accept that $0^n$ equals $\underbrace{0\times\cdots\times0}_{n\text{ times}}=0$, but when the exponent is any real number, I feel lost.

I have been looking into Rudin's book, but it doesn't seem like he has any use for proving/defining this or assumes it is trivial maybe. I don't know where else to look.

 

[martinbn]

For rational numbers you can accept that $0^{\frac pq} = \sqrt[q]{0^p} = 0$, and then by continuity for all positive reals.

 

[pasmith]

I suspect that assuming 0^{1/p} \neq 0 leads to a contradiction with the field axioms.

 

[psie]

I suspect that assuming 0^{1/p} \neq 0 leads to a contradiction with the field axioms.

I was hoping for that too, but I don't think the field axioms relate to exponentiation.

 

[fresh_42]

You will have to define what e.g. $a^\sqrt{2}$ means before you use it. The German Wikipedia follows the method @martinbn has suggested in post #2 as $a^r=\displaystyle{\lim_{n \to \infty}a^{q_n}}$ where $(q_n)\subseteq \mathbb{Q}$ is a sequence of rational numbers with $\displaystyle{\lim_{n \to \infty}q_n}=r.$

They also offer the alternative of
$$
a^r=\exp(r\cdot \log(a))
$$
for non-negative basis $a\ge 0.$ This uses the continuity of $\exp$ and $\log.$ For $a=0$ we still have to make a limit consideration and written it sloppy yields $0^r=\exp(-\infty )=0.$

 

[psie]

You will have to define what e.g. $a^\sqrt{2}$ means before you use it.

I agree. I was mainly looking at Rudin and I don't think he does. I don't know if he takes it for granted or just has no need for it in the book.

The German Wikipedia follows the method @martinbn has suggested in post #2 as $a^r=\displaystyle{\lim_{n \to \infty}a^{q_n}}$ where $(q_n)\subseteq \mathbb{Q}$ is a sequence of rational numbers with $\displaystyle{\lim_{n \to \infty}q_n}=r.$

They also offer the alternative of
$$
a^r=\exp(r\cdot \log(a))
$$
for non-negative basis $a\ge 0.$ This uses the continuity of $\exp$ and $\log.$ For $a=0$ we still have to make a limit consideration and written it sloppy yields $0^r=\exp(-\infty )=0.$

If we insist on exponentiation being continuous, then indeed $0^r=0$ for $r>0$. But does this not have as a consequence that $0^0=0$ also?

 

[fresh_42]

If we insist on exponentiation being continuous, then indeed $0^r=0$ for $r>0$. But does this not have as a consequence that $0^0=0$ also?

$0^0$ would mean $0^0=\exp(0\cdot \log (0))$ which is not defined at first glance. Hence we need to write $$0^0=\displaystyle{\lim_{n \to \infty}\exp(0\cdot \log (q_n))}=\exp(0)=1$$ with a sequence $(q_n)\subseteq \mathbb{Q}$ that converges to $0.$

This makes sense, as $0^0$ is an empty product and as such algebraically defined to be the neutral element in the multiplicative group, which is $1.$ The same as empty sums are defined to be $0.$ To define $0^0=0$ isn't an option, simply because $0$ is not part of the multiplicative group. However, sometimes it is convenient to set $0^0=0,$ IIRC e.g. in computer science, but this is an unusual exception in order to deal with extremes in specific formulas. In my opinion, $0^0=1$ should always be used since it is the only definition that makes mathematically sense.

 

[Infrared]

I agree. I was mainly looking at Rudin and I don't think he does. I don't know if he takes it for granted or just has no need for it in the book.

Chapter 1 exercise 6.

 

[fresh_42]

Here is another way to solve
$$
0^0=(1-1)^0=\sum_{k=0}^0 \binom{0}{k}1^k(-1)^{0-k}=\binom{0}{0}1^0(-1)^{0-0}=1.
$$

 

[Gavran]

There are only postulates those define zero raised to the power of a real number. See https://en.wikipedia.org/wiki/Exponentiation#Powers_of_zero.

 

[WWGD]

Well, if you accept $f(x)=0^x$is continuous at the Rationals ; $0^{p/q}= (0^p)^{1/q}=0^{1/q}$ and that the set of points of continuity of Real function is a $G_{\delta}$ set, then you would need to find a set $S$ containing $\mathbb Q^{+}$, with $S$ being a $G_{\delta}$ set. That $S$ must be the entire Positive Real line. You can use, e.g., $\Cap_{n=1}^{\infty} (1/n, \infty)$

 

[Gavran]

For example, it can be proved that $$ \lim_{x\to0^+}x^x=1 $$ but it can not be proved that $0^0=1$.
$0^0$ can only be defined by a convention. See https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero.

 

[fresh_42]

For example, it can be proved that $$ \lim_{x\to0^+}x^x=1 $$ but it can not be proved that $0^0=1$.
$0^0$ can only be defined by a convention. See https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero.

A convention that makes sense for many reasons. I have already mentioned at least three of them! The most important one is in my opinion the empty product. How do you define the notation $\text{anything}^0$?

This is the crucial question. Without answering it, the debate is meaningless. However, if you decide to define it as an empty product, then there is no alternative to $1$ simply because the multiplicative group $\left(\{1\},{\cdot}\right)$ does not contain any other element.

If you decide to define it by the base as $0^0=(1-1)^0=\ldots=1$ then you run into the binomial formula which also results in $1.$

If you decide to define it by analytical means, then you run into at least two continuity problems which have also been discussed here in multiple posts. Every setting besides $1$ causes problems. I wouldn't call that a convention. It's a natural result of various principles (continuity, binomial formula, empty product) and we usually don't want to lose either of them.

Wikipedia isn't flawless!