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So I encountered this definition in Salas Hille Etgen's One and Several Variables Calculus:

(This definition is for single variable case)

Definition: Let $g:\Bbb R \to \Bbb R$ be a function defined at least in some neighbourhood of $0$. We say that $g(h)$ is $o(h)$ and write $g(h)=o(h)$ to indicate that $\lim_{h\to 0}\frac{g(h)}{h}=0$.

So, after this definition, our class soon introduced a theorem:

Theorem: The following are equivalent:

$1. f(x+h)-f(x)-f'(x)h=o(h)$

$2. f(x+h)-f(x)=f'(x)h+o(h)$

And my professor gave us such proof:

$f(x+h)-f(x)=[f(x+h)-f(x)-f'(x)h]+f'(x)h=o(h)+f'(x)h$

So I began to doubt: What makes it legal to operate $o(h)$ as a normal function? Although it seems so natural, we still added something wasn't in the original definition. It is not an equation, it is a sentence, isn't it? Since the original definition write $g(h)=o(h)$ for merely implying $\lim_{h\to 0}\frac{g(h)}{h}=0$, is it okay to treat $o(h)$ like another number or function(e.g. replace $g(h)$ with $o(h)$ in the proof)?

Just like the definition of the notation of limit, it doesn't imply that it is a number. It is a sentence(description) replaced by abbreviations through the definition.

Hence, my question are summed up in 2:

  1. Is it legal to operate $o(h)$ in such manner?
  2. If not, is there another way to go through this proof?

Thanks in advance.

李智修
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  • The definition makes no sense to me. What is $h$? – Git Gud Aug 19 '18 at 09:26
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    in 2., did you mean to write f'(x) or f'(x)h? – Calvin Khor Aug 19 '18 at 09:27
  • OK, for anyone wondering, the proper definition goes like this. Let $g\colon \mathbb R\to \mathbb R$ be a function defined in an open-interval around $0$, we say that "$g$ is $o(\mathrm{id})$" and write $g=o(\mathrm{id})$ if, and only if, $\lim \limits_{x\to 0}\left(\dfrac{g(x)}{\mathrm{id}(x)}\right)=0$, where $\mathrm{id}$ denotes the identity function. – Git Gud Aug 19 '18 at 09:33
  • @GitGud don't think I understand the distinction, and I'd like to see an actual piece of mathematics written with $o(\operatorname{id})$. I think its too clunky a notation. Perhaps you'd be happy with $g(x)=o(x)$ as $x\to 0$? – Calvin Khor Aug 19 '18 at 09:36
  • @CalvinKhor You won't find mathematics written like this because people who care more about the notation than analysts, aren't analysts, I think you do understand the distinction, perhaps you meant to say that you don't see how it is useful? It's useful for me for the simple fact that I couldn't understand the definition in the question and also for the fact that the definition I wrote, however unintelligible it might seem to many, is actually correct. This is important because when something is incorrect, you're left with guessing. If something is correct, you have hope of understanding it. – Git Gud Aug 19 '18 at 12:29
  • I wouldn't be happy with your suggestion because the concept of little-oh is tied to neighborhoods of $0$. It's important for me to make a distinction between variables and functions, $x$ isn't a function, $x\mapsto x$ is. – Git Gud Aug 19 '18 at 12:32
  • I just realised what most troubles me about the definition in the question. In "$g(x)=o(x)$", the symbol $x$ is being used both as a variable and as a function. Sorry, even if you're notationally abuse-prone, you can't have both. – Git Gud Aug 19 '18 at 12:35
  • @GitGud I don't comprehend at all the second point, it seems you are also against the notation $1/n \to 0$ as $n\to \infty$. Re the first, I actually really feel like I don't understand the difference because I already understood the commonly taught meaning of what was written above to be what you wrote, up to change/abuse of notation. I'll grant that a martian may not understand the mathematics as written... – Calvin Khor Aug 19 '18 at 12:50
  • @GitGud I do remember struggling a lot with coming to terms with this notation and others as I try to be an analyst. I think I try to accept it on the grounds of practicality, eg $\text{id}^2$ isn't what I want it to mean, and I don't want to write my multivariate polynomials as $\text{id}_1^3\text{id}_2^4+...$. Perhaps if we are okay with $x$ as a polynomial distinguished from the function given by evaluation of a polynomial, we can have $o(\cdot)$ take these polynomials (and extensions eg $e^x$)? – Calvin Khor Aug 19 '18 at 13:10
  • @CalvinKhor Yup, $f'(x)$ was a typo, I meant $f'(x)h$, thank you :). – 李智修 Aug 19 '18 at 14:19
  • @GitGud $h$ is simply a small number(I guess? since the text book didn't mentioned it). – 李智修 Aug 19 '18 at 14:23
  • @CalvinKhor My issue in the second point wasn't with the arrow sign, per se, but rather specifying that we're talking about the origin. By definition little-oh is about the origin. In the case of polynomials, if you mean actual polynomials, as formal expressions, then using $x$ is perfectly correct. But if you mean polynomial functions, then my preference is to say something like, "let $f(x, y, z)=x^2y+y^2z+3$, for all real numbers $x, y$ and $z$" circumventing completely the (read my) necessaity of writing $\mathrm{id}_{\mathbb R}$ or $t\mapsto 3$. – Git Gud Aug 19 '18 at 15:22
  • @李智修 Nope, $h$ is being used both as a function, namely the identity function as a number. I didn't realise it at first, but that's what I find so confusing about this definition. – Git Gud Aug 19 '18 at 15:23
  • @GitGud I was taught that $f(x) = o(g(x))$ as $x\to x_0$ in $S$ if $\lim_{x\to x_0} \frac{f(x)}{g(x)} = 0 $ as $x \to x_0$ in (the induced subspace) $S$, which is more general. e.g. $e^{-z^2} = o(1/z)$ as $z\to\infty\in\overline{\mathbb C}$ in ${ z \in \mathbb C : 0< \arg z < \pi/100 }$. I don't know how to voice my thoughts better, but perhaps a good way to think about the abuse is to have the formal expressions $\sin\text{x} = o(\text{x})$, defined using the extension of the evaluation map we know from polynomials to polynomial functions. Thank you very much for your time and discussion. – Calvin Khor Aug 19 '18 at 15:59

2 Answers2

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I know that writing $g(h)=\mathrm o(h)$ is a thing, but I think it is one of the most misleading notations ever $-$ when used without further explanation. So I will try to give the explanations: define $\mathrm o(h)$ as a set of functions:

$$\mathrm o(h):=\left\{f:\Bbb R\to\Bbb R \mid \lim_{h\to 0} \frac{f(h)}h = 0\right\},$$

and then $g\in \mathrm o(h)$ is what formerly was denoted by $g(h)=\mathrm o(h)$. Now we add a notational convention: any expression that contains $\mathrm o(h)$ as if it where a function is to be interpreted as "there is a function $g\in\mathrm o(h)$ replacing this occurence of $\mathrm o(h)$". So e.g. $f(x+h)-f(x)-f'(x) = \mathrm o(h)$ means

$$\forall x:\exists g_x\in\mathrm o(h):f(x+h)-f(x)-f'(x)=g_x(h).$$

Note that this might be a different $g_x$ for every $x$ (the derivative is defined point-wise). Now it might be clear why we can handle this "function" $\mathrm o(h)$ as if it where a real function. The right side above is obviously equivalent to

$$\forall x:\exists g_x\in\mathrm o(h): f(x+h)-f(x)=f'(x)+g_x(h).$$

And by our notational convention, we can write this as $f(x+h)-f(x)=f'(x)+\mathrm o(h)$.

M. Winter
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  • Actually $\exists$ can be replaced by $\exists !$, right? –  Aug 19 '18 at 10:12
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    @user529760 Yes, or we can write $\forall x: f(x+h)-f(x)-f'(x)\in\mathrm o(h)$. But then bringing $f'(x)$ to the other side again needs some explanations. – M. Winter Aug 19 '18 at 10:13
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    Point 10 here. – Git Gud Aug 19 '18 at 12:33
  • I can handle some abuse of notation, but $g(h)=o(h)$ really is disgusting. It pleases the eye to see $g\in o(h)$ even if I don't entirely agree with this notation. Upvoted. – Git Gud Aug 19 '18 at 12:38
  • I also write $f\in o(h)$ rather than use equality. I haven't checked it, but perhaps in terms of equivalence classes $f$ can be identified with $o(h)$, justifying the equals notation. – AlvinL Aug 20 '18 at 08:19
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I believe this is your definition (assuming, some definition of $o(\cdot)$) $$ a = b + o(c) \text{ iff there exists } d \text{ such that } d=o(c), \text{ and } a-b=d$$

So the proof- let $x$ be fixed. Then note that we can always define $R(h)$ to force the following two equalities at the same time, $$f(x+h) - f(x) - f'(x) h = R(h) \iff f(x+h) - f(x) = f'(x) h+ R(h)$$ Then statement 1. makes the claim that in the LHS equation, $R(h) = o(h)$. The statement 2. (assuming you meant f'(x)h and not f'(x)) asserts again that $R(h) = o(h)$ but starting from the RHS. That these are equivalent is the above $\iff$ arrow.

Calvin Khor
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