Isn't $$f(z+h)=f(z)+f'(z)h+o(h)$$ an abuse of notation ?

Indeed, this means that there is a function $\varepsilon $ s.t. $\varepsilon (h)\to 0$ when $h\to 0$ s.t. $$f(z+h)=f(z)+f'(z)h+h\varepsilon (h).$$

But I guess that if $\bar z\neq z$, and if $$f(\bar z+h)= f(\bar z)+f'(\bar z)h+o(h),$$ then $$f(\bar z+h)=f(\bar z)+f'(\bar z)h+h\varepsilon (h),$$ will not be true, so the function $\varepsilon $ depend on $z$. Shouldn't it be more clear to write $$f(z+h)=f(z)+f'(z)h+o_z(h),$$ to mention that the function $\varepsilon $ will depend on $z$ ?

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    $\begingroup$ Your observation is correct. Your reasoning about clarity is like saying that every statement in the form $\forall a,\exists b,\cdots$ should rather be written as $\forall a,\exists b_a,\cdots$. $\endgroup$ – Gae. S. Aug 14 at 14:07
  • $\begingroup$ To add to this, you observed that there is an implicit dependence on $z$ and whether this constitutes an abuse of notation is entirely subjective. Does it successfully communicate what is intended? Probably. If uniform differentiability were intended (as in the most literal interpretation) then I’d find it very confusing if uniformity it weren’t explicitly stated. $\endgroup$ – adfriedman Aug 14 at 19:57

Using $=$ with little-$o$ is the abuse of notation that happens here. And one technically ought to specify $o(h)$ as $h\to0$.

The function $\varepsilon$ will depend on $z$, but $o(h)$ is a class of functions. If $f$ is differentiable, then $h\varepsilon_z(h)$ will fall under this class, and vice versa. So the $o$ is the same from $z$-value to $z$-value, and therefore we don't really need $o_z$.

  • $\begingroup$ nice answer. I get it :) So, my other abuse of notation was : there is function $\varepsilon $ such that... and should have been there is a function $\varepsilon _z$, right ? (or to say there is a function $\varepsilon $ s.t. $\varepsilon (z,h)\to 0$ when $h\to 0$). At the end, we rather write $\varepsilon \in o(h)$ than $\varepsilon (h)=o(h)$, no ? $\endgroup$ – John Aug 14 at 14:12
  • $\begingroup$ @John Yes, one could, for clarity, say "for any $z$ there is a number $f'(z)$ and a function $\varepsilon_z$ such that ..." I have seen $\epsilon$-$\delta$ proofs use $\delta_{x,\epsilon}$ for clarity as well. But it is not necessary. No two $\varepsilon$s will ever meet, so there isn't really any ambiguity. And I do prefer $\in$, yes. Unfortunately, $=$ is too widespread to really do anything about it. $\endgroup$ – Arthur Aug 14 at 14:22

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