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is plugging $0$ in (6) result to $0^0$?

here is conditions of $8.1$

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Mathematics enjoyer
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1 Answers1

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Yes.

Rudin is using the convention that $0^0 = 1$.

There are a lot of arguments in favour of this convention, but really in this case it just comes down to: "if I use the convention that $0^0 = 1$, then all my formulas hold well, whereas if I use a different convention, then I'll have to separate the $n=0$ term from the rest of the sum everywhere throughout my book, resulting in heavier, harder to read formulas."

Also note the two related conventions that Rudin might also be using: \begin{align*} 0! & = 1 \\ \prod_{n \in \emptyset} c_n & = 1 \end{align*}

Stef
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  • but this doesn't seem to be rigorous at all! – Mathematics enjoyer Dec 25 '23 at 14:08
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    @Mathematicsenjoyer What's not rigorous? Defining a convenient notation? – Stef Dec 25 '23 at 14:09
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    @Mathematicsenjoyer Note that it's okay to define whatever notation you want. The one thing you're not allowed to do is use the same notation for two different things and draw abusive conclusions by equaling the two things just because they have the same notation. – Stef Dec 25 '23 at 14:13
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    @Mathematicsenjoyer As an example, remember how at some point a math teacher defined $a^n = a \times a \times ... \times a$ for any real $a$ and positive integer $n$, and then later another math teacher defined the exponential function $\exp(x) = \sum x^k / k!$, and then the teacher wants to define the notation $e^x = \exp(x)$. But at this point $e^n$ would be an ambiguous notation: it might mean either $e \times ... \times e$, or $\exp(n)$. So first, the teacher has to prove that $\exp(n) = e \times ... \times e$ when $n$ is integer; and then it's okay. – Stef Dec 25 '23 at 14:16
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    @Mathematicsenjoyer So here it's the same thing; Rudin defines $0^0$ as a notation which is equal to $1$; and before choosing this convention, he made sure that it wouldn't lead him to draw abusive conclusions. – Stef Dec 25 '23 at 14:17
  • BUT $0^0 = \frac{0}{0}=1??$ – Mathematics enjoyer Dec 25 '23 at 14:18
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    @Mathematicsenjoyer No. $\frac 0 0$ does not make sense; I'm not aware of any mathematician who ever used the weird, confusing, useless and error-prone convention that $\frac 0 0 = 1$. However $0^0 = 1$ is used a lot and there are good reasons for that. In fact there is more that one question about it on this website, with insightful answers as to why it is a good convention, and also with mentions about the (small) limitations and caveats of this notation. – Stef Dec 25 '23 at 14:20
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    I meant that $0^0 = 0/0$ since $a^{b-1}=\frac{a^b}{a}$ – Mathematics enjoyer Dec 25 '23 at 14:23
  • since $0^0=1$ then $0/0 =1$!!? – Mathematics enjoyer Dec 25 '23 at 14:24
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    @Mathematicsenjoyer $a^{b-1} = \frac{a^b}{a}$ is true for every real $b$ and non-zero real $a$. No one ever said it was true for $a = 0$, so there is no contradiction or paradox. Be careful when you make statements like $a^{b-1} = \frac{a^b}{a}$ without saying what is $a$ and $b$. It's a bit like saying "$a^2 + b^2 = c^2$" without saying "if $a$, $b$ and $c$ are the lengths of the three sides of a right triangle, with $c$ being the length of the hypotenuse" – Stef Dec 25 '23 at 14:26
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    in the complex world is the function $x^x$ is continuous at 0 by defining $0^0=1$? – Mathematics enjoyer Dec 25 '23 at 14:28
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    @Mathematicsenjoyer The real function $f : \mathbb R \to \mathbb R$ defined by $f : x \mapsto x^0$, with the convention $0^0 = 1$, is continuous. The real function $g : [0, +\infty) \to \mathbb R$ defined by $g : x \mapsto 0^x$, with the convention $0^0 = 1$, is not continuous at 0, since $g(x) = 0$ for all $x > 0$ and yet $g(0) = 1$. The real function $h : [0, +\infty) \to \mathbb R$ defined by $h : x \mapsto x^x$, with the convention $0^0 = 1$, is continuous, and here is a proof. – Stef Dec 25 '23 at 14:35
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    @Mathematicsenjoyer Defining $x^x$ for a negative real $x$ or for a complex $x$, raises a lot more questions than just defining $0^0$, so I won't get into this in these comments. See for instance this related question. In particular, defining $\log(z)$ for complex $z$ raises a lot of questions, and defining $a^b$ for $a$ and $b$ complex raises a lot of questions, and no matter your definition you won't be able to keep all the theorems you know about $a^b$ when $a$ and $b$ are real. – Stef Dec 25 '23 at 14:39
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    @Mathematicsenjoyer For instance, you already know that for any real numbers $a, b$ and $c$, we have: $a^{b+c} = a^ba^c$. Now if you define $a^b$ for complex numbers, you'll have to remember one of my earliest comments, I said "it's okay to define whatever notation you want. The one thing you're not allowed to do is use the same notation for two different things and draw abusive conclusions by equaling the two things just because they have the same notation." – Stef Dec 25 '23 at 14:41
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    @Mathematicsenjoyer This warning applies as well when extending a notation, so you have to prove your theorems again; you can't just conclude $a^{b+c} = a^ba^c$ for your new notation without proving it again. Previously you had only proven it for real numbers, so when you give a definition of $a^b$ for complex numbers you cannot conclude that $a^{b+c} = a^ba^c$ for complex numbers. – Stef Dec 25 '23 at 14:42
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    @Mathematicsenjoyer For all these reasons, when making new definitions, mathematicians tend to use notations that "work well": if you use a notation that looks similar to an already-existing notation, you have better be very careful and make sure that this similarity does not result in ambiguity and mistakes. – Stef Dec 25 '23 at 14:47
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    Thank you very much sir for you time and effort. – Mathematics enjoyer Dec 25 '23 at 14:50