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I am reading Walter Rudin's "Principles of Mathematical Analysis".

There is the following theorem in this book:

p.57
Theorem 3.20(a)
If $p > 0$, then $\lim_{n\to\infty} \frac{1}{n^p}=0$.

Take $n > (\frac{1}{\epsilon})^{\frac{1}{p}}$. Then $n^p > ((\frac{1}{\epsilon})^{\frac{1}{p}})^p = \frac{1}{\epsilon}$. So $\epsilon > \frac{1}{n^p}$. To prove $((\frac{1}{\epsilon})^{\frac{1}{p}})^p = \frac{1}{\epsilon}$, I think we need the property $(a^x)^y = a^{x y}$. And Rudin didn't write this property on p.22 ex6.

On p.22 Exercise 6, Rudin defined $b^x$ for $b \in \{y \in \mathbb{R} | y > 1\}, x\in\mathbb{R}$.
And the reader proves that $b^{x+y} = b^x b^y$ for all $x, y \in \mathbb{R}$.

But Rudin didn't define $b^x$ for $b \in \{y \in \mathbb{R} | 0 < y \leq 1\}, x\in\mathbb{R}$.

And Rudin didn't write other properties of $b^x$.
For example, Rudin didn't write $(b^x)^y = b^{xy}$ for all $x, y \in \mathbb{R}$.

I am disappointed and sad.
Walter Rudin's "Principles of Mathematical Analysis" isn't perfect.

I can guess $b^x$ is defined as $(\frac{1}{b})^{-x}$ for $b \in \{y \in \mathbb{R} | 0 < y \leq 1\}, x\in\mathbb{R}$.

Isn't Walter Rudin's "Principles of Mathematical Analysis" self-contained?

Is there a self-contained analysis book in the world?

tchappy ha
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    You are making conclusion too hastily. $x^\alpha$ for positive $x$ and real $\alpha$ is defined on p.181. That said, most reviewers on the internet seem to agree that Rudin is not suitable for self studies (if one learns analysis for the first time) and it should be used as a textbook in a first course on analysis only if the students are guided by a good lecturer. – user1551 Jan 21 '19 at 13:09
  • Thank you very much for the information, user1551. Now I am worried about a circular argument because Rudin uses $x^\alpha$ before he defines $x^\alpha$. – tchappy ha Jan 22 '19 at 01:07
  • Thank you very much again, user1551. Rudin didn't write $(a^x)^y = a^{x y}$ on p.22 ex.6. But we must use this property in the proof of Theorem 3.20(a) on p.58. – tchappy ha Jan 22 '19 at 01:54
  • Thank you very much again, user1551. Take $n > (\frac{1}{\epsilon})^{\frac{1}{p}}$. Then $n^p > ((\frac{1}{\epsilon})^{\frac{1}{p}})^p = \frac{1}{\epsilon}$. So $\epsilon < \frac{1}{n^p}$. To prove $((\frac{1}{\epsilon})^{\frac{1}{p}})^p = \frac{1}{\epsilon}$, I think we need the property $(a^x)^y = a^{x y}$. And Rudin didn't write this property on p.22 ex6. – tchappy ha Jan 22 '19 at 03:24
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    Even if you want to use $(a^x)^y=a^{xy}$, there still isn't a circular argument, because the general definition of exponential on p.181 does not depend on the result of the 3.20(a). And strictly speaking, you don't actually need $(a^x)^y=a^{xy}$, because there are other ways to prove the statement. E.g. just pick a sufficiently large $n$ such that $n>(\frac1\epsilon)^{1/q}$ for some rational number $0<q\le p$ instead. Then $\frac1{n^p}\le\frac1{n^q}<\epsilon$. – user1551 Jan 22 '19 at 04:20
  • Thank you very much for your answer, user1551. I am relieved now. – tchappy ha Jan 22 '19 at 04:50
  • +1 for the sentiments expressed in last sentence. In general writing a self contained maths textbook is difficult. Especially those books which are used as standard texts in classroom setting are never meant to be like that. – Paramanand Singh Jan 22 '19 at 08:48
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    A simpler approach is to consider a positive integer $k>1/p$ and then $1/n^p<1/n^{1/k}$ and this is less than $\epsilon$ if $n>\epsilon^{-k}$. – Paramanand Singh Jan 22 '19 at 08:54
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    Also the approach by Rudin to define $b^x$ is somewhat complicated. A better approach is to develop logarithms first. Or if you want to avoid logarithm then better use limits instead of sup or inf. Limits obey nice algebraic properties which inf and sup may not. See this post for more details. – Paramanand Singh Jan 22 '19 at 08:57
  • Thank you very much for the simpler approach, Paramanand Singh. In Michael Spivak's "Calculus", I found a definition of $\exp(x)$ as the inverse function of $\int_{1}^x 1/t dt$. I guess this definition is "a better approach" you are saying. This definition is very simple and I like it. Thank you. – tchappy ha Jan 22 '19 at 09:35
  • Thank you very much, Paramanand Singh for your blog. I will read your post now. – tchappy ha Jan 22 '19 at 09:40
  • @ParamanandSingh I am reading your blog. I think we must prove $\lim_{n\to\infty} (1 + \frac{x}{n})^n$ exists first to prove (8). For example, $\lim_{n\to\infty} \sin(\pi n) = 0$, but $\sin(\lim_{n \to \infty} \pi n)$ diverges. – tchappy ha Jan 22 '19 at 12:44
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    @tchappyha: you have caught a minor (but worth noting) issue here! If the function $f$ is continuous and strictly monotone then we can exchange $f$ and limit operation without worrying about the existence of limit. See the theorem mentioned at the end of this answer: https://math.stackexchange.com/a/1073047/72031 I will update this in blog after some time. – Paramanand Singh Jan 22 '19 at 14:19
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    Another way out is to just show that $n\log(1+(x/n))\to x$ and hence applying continuous $\exp$ we get $(1+(x/n))^n=\exp (n\log(1+(x/n)))\to \exp(x) $. – Paramanand Singh Jan 22 '19 at 14:26
  • @ParamanandSingh I understood the second way, but I wanna understand the first way too. So, I try to understand the first way tomorrow. Thank you very much. – tchappy ha Jan 22 '19 at 15:26
  • @ParamanandSingh Rudin defined $b^x$ using supremum probably because it is a natural extension of fractional exponents. He later did define the exponential function $\exp(x)$ (on pp.180-181) and show that $\exp(x)$ is consistent with $e^x$. Therefore, he successfully bridged the gap between high school mathematics (where $b^x$ is not rigorously defined but is introduced along the line of fractional exponents) and undergraduate real analysis... – user1551 Jan 22 '19 at 18:53
  • ... and I believe this is actually where Baby Rudin shines: the choice and arrangement of material are really thoughtful (at least in the first eight chapters). But Baby Rudin also sucks (I'm sorry if the language is offensive) because it has almost no contexts or expositions. While there are good reasons behind his arrangement of material, Rudin doesn't tell the readers those reasons and his readers are left clueless. – user1551 Jan 22 '19 at 18:53
  • @ParamanandSingh If $f$ is a continuous and strictly monotone function, then $f$ has the inverse function $f^{-1}$. If $\lim_{n\to\infty} f(a_n)$ exists, then $\lim_{n\to\infty} f^{-1}(f(a_n)) = \lim_{n\to\infty} a_n$ exists because $f^{-1}$ is also continuous. Thank you very much for your answer. – tchappy ha Jan 22 '19 at 23:44
  • @user1551: IMHO a textbook should at least give some motivation behind the presentation of the material apart from ensuring that the presentation is easy to understand (for the intended audience). – Paramanand Singh Jan 23 '19 at 06:54

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You can easily extend the definition of $b^x$ for $0 < b \leq 1$ by defining $$b^x := \frac{1}{(1/b)^x}$$ and show the property $$b^{x+y} = b^x b^y$$ still holds.

A. Bailleul
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