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Let's define $o(g(x))$ as usually:

$$ \forall x \ne a.g(x) \ne 0 \\ f(x) = o(g(x)) \space \text{when} \space x \to a \iff \lim_{x \to a} \frac{f(x)}{g(x)}=0 $$

By Taylor expansion of $e^u$ at $0$, we have

$\tag{1} e^u = 1 + u + o(u) \text{ when } u \to 0 \iff \lim_{u \to 0} \frac{e^u - 1 - u}{u} = 0$

How does the little-o look like when I use the above Taylor expansion at $g(x)$, where $g(x) \to 0 \text{ as } x \to 0$?

If I just replace $u$ with $g(x)$ in $(1)$, I get:

$e^{g(x)} = 1 + g(x) + o(g(x))$.

But how to interpret the $o(g(x))$ there?

It is a function which is of a lower order than $g(x)$, as $g(x) \to 0$, but I'm not sure how that fits into the little-o definition above, which does not say that $g(x) \to a$, but $x \to a$.

Note: This question is related to this question, but still different (at least as far as I can see), because the function here is not a complete composition, because we do not have $f(g(x))$ in the numerator, but only the denominator changes.

S11n
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1 Answers1

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You can do this expansion as is if $g(x) \to 0$ as $x \to 0$. If this is not the case then the way you are expanding $\exp$ is already not appropriate, and you should instead be considering the behavior of $\exp$ near $g(0)$ (assuming $g$ is continuous).

The general way to make the adjustment looks like

$$f(g(x))=f(g(0)+g(x)-g(0))=f(g(0))+f'(g(0))(g(x)-g(0))+o((g(x)-g(0))$$

as $x \to 0$. Note how the leading terms now include a $-f'(g(0)) g(0)$ term that was not there in your formulation, which now ensures that the leading terms approach the desired function as $x \to 0$ whether $g(0)=0$ or not.

Ian
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  • What do you mean by "not appropriate"? Shouldn't just the error term be larger, but the equation $(1)$ remains? I would appreciate if you could expand the answer, so it is clear to me how to treat the remainder when we do the $T_1 e^u(f(x); 0)$. – S11n Jun 06 '22 at 09:47
  • @S11n The expansion you did in the first place requires the argument of $\exp$ to be going to zero. It isn't doing that in the general situation of a composition, so you have to do something else. – Ian Jun 06 '22 at 10:03
  • Thank you! It is now clear to me that the little-$o$ formulation I wrote in $(1)$ is valid only when the point $x$ at which $T_1 e^u(x; 0)$ is computed tends to $0$, so I should delete the last part of the last sentence in my question. However, the main question I had remains, and that is how to interpret the $o$ term there, when we do not have a point, but if we have a function (which tends to $a$ as $x \to a$, or $g(x) \to 0$ in that concrete case as $x \to 0$). – S11n Jun 08 '22 at 07:52
  • @S11n The little-oh notation itself doesn't demand that $g$ goes to zero in the relevant limit. A function $h$ in $o(g(x))$ as $x \to a$ still by definition has $\lim_{x \to a} \frac{h(x)}{g(x)}$. What demands that $g$ actually goes to zero in this situation is the fact that you wanted to use an expansion of the outer function that holds when the argument of the outer function is close to $0$. As an example to illustrate that, if $g(x)=x$ and $a=1$ then $\lim_{x \to 1} \frac{e^x-1-x}{x}$ is not zero, but rather $e-2$. – Ian Jun 08 '22 at 09:32
  • Thank you Ian, but unfortunately it seems that I do not explain my confusion well. I understand that if I would just see $o(g(x))$ as $x \to 0$, then what you say is correct. But when I see $o(g(x))$ as $g(x) \to 0$ (which is the case here, because I just replaced $u$ with $g(x)$), then I don't understand why is that obviously the same thing as $o(g(x))$ as $x \to 0$. – S11n Jun 08 '22 at 17:20
  • @S11n It isn't, unless $g(x) \to 0$ as $x \to 0$. That's why it doesn't make sense to do that same expansion on $\exp$. (Not only does it not make logical sense but the actual desired result just doesn't hold; $e^x-1-x$ is not $o(x)$ as $x \to 1$.) You have to expand $\exp(u)$ around $u=g(0)$ in order to expand $e^{g(x)}$ around $x=0$. – Ian Jun 08 '22 at 17:21
  • yes, and that's exactly what I currently don't know how to prove (that that implication holds): $g(x) \to 0 \text{ as } x \to 0 \land f(x) = o(g(x)) \text{ as } g(x) \to 0 \implies f(x) = o(g(x)) \text{ as } x \to 0 $ – S11n Jun 08 '22 at 17:23
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    @S11n You want to show that for any $\epsilon_1>0$ there exists $\delta_1>0$ such that $\frac{|f(x)|}{|g(x)|}<\epsilon$ when $|x|<\delta_1$. You know that for any $\epsilon_2>0$ there exists $\delta_2>0$ such that $|g(x)|<\epsilon_2$ when $|x|<\delta_2$. And you know that for any $\epsilon_3>0$ there exists $\delta_3>0$ such that if $|g(x)|<\delta_3$ then $\frac{|f(x)|}{|g(x)|}<\epsilon_3$. – Ian Jun 08 '22 at 17:30
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    @S11n (Cont.) So now you're given $\epsilon_1$. Set $\epsilon_3=\epsilon_1$ to get a $\delta_3$ that sets how small $g$ needs to be. Now invoke the second statement with $\epsilon_2=\delta_3$ to get a $\delta_2$ that determines how small $x$ needs to be. Then finally set $\delta_1=\delta_2$ to finish the proof. – Ian Jun 08 '22 at 17:30
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    Sorry to say that I'm not sure what a clean, no epsilon-delta way of deriving rigorously would look like. The intuition is basically the same as with $\lim_{x \to a} f(g(x))=f \left ( \lim_{x \to a} g(x) \right )$ for continuous $f$ (pass the $\delta$ from the outer function in as the $\epsilon$ in the inner function), but it's not really the same result. – Ian Jun 08 '22 at 17:32
  • About one of the comments above, as $x\to 0$, we would have $g(x)\to 0$, so (since $f(x)=o(g(x))$ as $g(x)\to 0$), we have $$\lim_{x\to 0}\frac{f(x)}{g(x)}=0.$$ But all of this would show, $f(x)=o(g(x))$ as $x\to 0$. – Julian Mejia Jun 08 '22 at 17:35
  • @Ian Yes, when you frame the question via the limit definition, then the proof is obvious. Thanks! It seems you already spent a lot of time explaining this to me, and it would be negligible to just add those steps to your answer so it makes sense with the question. Otherwise, I would just delete the question, because people don't seem to like it :) – S11n Jun 09 '22 at 16:09
  • @JulianMejia I added a note to the question, so it is clear why this question is different from the question of the function composition, which seems that you are assuming in your short answer, but Ian didn't miss in his last comment (when saying that the two are not quite the same). – S11n Jun 09 '22 at 16:18
  • @Ian Thanks! I also understood why the two are not the same results (as I mentioned in the note in the question I've just added, after you hinted on that). – S11n Jun 09 '22 at 16:20