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Question:

Is algebra on equivalent infinitesimals valid?

Example:

$$\sqrt{1+x} - 1 \sim_0 \frac{x}{2} \longrightarrow \sqrt{1+x} \sim_0 \frac{x}{2} + 1$$

I'm aware that both the left and right equalities are true, but is this just a coincidence that it looks like I added $1$ to both sides of the left equality to get the the right?

Fine Man
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  • I recommend using something like $\sim_0$ to make it clear where the equivalence occurs :-) – Simply Beautiful Art Mar 08 '17 at 21:59
  • @SimplyBeautifulArt -- OK, sorry; I just editted it. The book I use is an old Soviet exercise book (Demidovich). I guess they had a shortage of ink so they omitted the $_0$s. :) – Fine Man Mar 08 '17 at 22:03
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    Well, $\sim_0$ is definitely closed under adding or multiplying by a constant. – Berci Mar 08 '17 at 22:07
  • @Berci -- I thought so, but I can only confirm using intuition. – Fine Man Mar 08 '17 at 22:08
  • Beware equivalence of functions is not compatible with addition in general. Counter example: $x+x^2\sim_0x$, $-x\sim_0-x$, but we can't deduce $x^2\sim_0 x-x=0$. – Bernard Mar 08 '17 at 22:34

3 Answers3

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We can be more rigorous using remainders:

$$\sqrt{1+x}=1+\frac x2+\mathcal O(x^2)$$

Where $\mathcal O(x^2)$ here means that as $x\to0$, then the remainder is at most $Cx^2$ for some $C$. Using Taylor polynomials and their remainders, we can see that for $|x|<0.5$,

$$\left|\sqrt{1+x}-1-\frac x2\right|\le2^{-2.5}x^2$$

Indeed, try this out and you will see it holds true, which allows you to show that

$$\sqrt{1+x}\sim_01+\frac x2$$

Anyways, in general, when you write something like

$$\sqrt{1+x}-1\sim_0\frac x2$$

It means that they both have the same growth rates (their difference is bounded by something that goes to zero faster than them) and from there it's easy to show addition and multiplication can be moved around.

Simply Beautiful Art
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  • The top half is a bit over my head. However, regarding the bottom half: are exponents also allowed (i.e. $\sqrt{1+x} \sim_0 1+\frac{x}{2} \longrightarrow 1+x \sim_0 (1 + \frac{x}{2})^2$)? – Fine Man Mar 08 '17 at 22:14
  • The basic idea is to remove the equivalence and replacing it with equality, holding some "remainder" and showing the remainder is bounded.$$f(x)\sim g(x)\implies R(x)=f(x)-g(x)\implies|R(x)|\le?$$From there, we would then have$$f(x)=g(x)+R(x)\implies[f(x)]^2=[g(x)+R(x)]^2$$So it is always the case that $[f(x)]^2\sim[g(x)]^2$. – Simply Beautiful Art Mar 08 '17 at 22:17
  • Or sometimes we use $R(x)=\frac{f(x)}{g(x)}$. It honestly depends on your situation. – Simply Beautiful Art Mar 08 '17 at 22:20
  • Equivalence of functions is perfectly rigourous! – Bernard Mar 08 '17 at 22:36
  • @Bernard Yes, it is indeed. – Simply Beautiful Art Mar 08 '17 at 22:43
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Adding is valid. "Equivalent infinitesimals" $f(x) \sim_0 g(x)$ are characterized by the fact that $f,g \to 0$ as $x \to 0$ and $\lim_{x \to 0} \frac{f(x)}{g(x)} = 1$. Using standard limit rules we can see that $h(x) := f(x) + c$ and $q(x) := g(x) + c$ satisfy $\lim_{x \to 0} \frac{h(x)}{q(x)} = 1$. Multiplication by a constant is also (trivially) valid: the common factor cancels out.

That said, in general, you should be careful when doing algebra to equivalents. Even the classic $\sin x \sim_0 x$ is not invulnerable, since $$\lim_{x \to 0} \frac{\exp \left(\frac{1}{x^2}\right)}{\exp\left(\frac{1}{\sin^2 x}\right)} \neq 1$$

which means $\exp \left(\frac{1}{x^2}\right)$ is not $\sim_0$ $\exp\left(\frac{1}{\sin^2 x}\right)$ which means you can't do whatever you want to equivalent infinitesimals and still maintain equivalence.

However, my hunch is that if $h$ is a sufficiently "well behaved" function, we have that $f \sim g$ implies $h\circ f \sim h \circ g$

I've posted a similar question here. In my question, the equivalents are not infinitesimal, but rather infinite, but the concept is similar.

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MathematicsStudent1122
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In some of the comments there was a bit of confusion whether the relation $a\sim_0 b$ is supposed to define an additive one, meaning that $a-b$ is infinitesimal, or a multiplicative one, meaning that $\frac{a}{b}$ is infinitely close to $1$, i.e., $1-\frac{a}{b}$ is infinitesimal. Not distinguishing between the two relations can certainly lead to error, as pointed out by user Bernard in a comment.

Therefore it would be safer to introduce distinct notation for the two relations. Thus, $a\approx b$ means that $a-b$ is infinitesimal, whereas $a\;_{\ulcorner\!\urcorner}\;b$ (similar to Leibniz's notation for equality) means that $1-\frac{a}{b}$ is infinitesimal, i.e., $1-\frac{a}{b}\approx0$.

If what you wish to use is the relation $a\;_{\ulcorner\!\urcorner}\;b$ then strictly speaking your deduction requires further justification since the relation is not preserved additively, but only multiplicatively.

For more details on Leibniz's relation see this answer.

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Mikhail Katz
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