0

It's true that $f(x+\Delta x)\approx f(x)+\Delta f$, but is it still correct that $f(x+\mathrm{d}x)=f(x)+\mathrm{d}f$ ? If so, what's the proof?

Since $\mathrm{d}x$ represents an infinitesimal quantity, this seems to be true.

This is related to the formal definition of $\mathrm{d}f$. If $\mathrm{d}f$ is defined as a differential form, it's a function $\mathbb{R}\rightarrow\mathbb{R}$, and it's also clear that $\mathrm{d}f=f'(x)\mathrm{d}x$.

The formal definition of $f'(x)$ is $$\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}~,$$ so $\mathrm{d}f$ is defined after $f'(x)$, and this looks rather circular to me.

Related links: What exactly is a differential? Rigorous definition of "differential"

nmasanta
  • 9,222
Ladmon Draxngfüskiii
  • 3
  • 1
    $df/dx$ is more of an operator, whereas $Δx$ is a specific number. – Korra Aug 26 '19 at 15:34
  • 5
    What are $\mathrm dx$ and $\mathrm df$? – Maximilian Janisch Aug 26 '19 at 15:36
  • Yes you are right.. but it would be better to right df as $f'(x)dx$. You can get the second equation in your question using definition of derivatives. – Korra Aug 26 '19 at 15:44
  • At the very least, you have to define what $dx$ and $df$ mean. Those are tricky concepts. – Thomas Andrews Aug 26 '19 at 15:57
  • What's the proof? You almost just proved it. If you let $\Delta x$ go to $0,$ then it becomes the differential $\mathrm d x.$ If $f$ is continuous, then $\Delta f$ also goes to $0$ and hence becomes differential. – Allawonder Aug 26 '19 at 16:00

2 Answers2

3

The answer is "yes and no", and it's that for multiple reasons. First of all the notations "$dx$" and "$df$" aren't super clear, and so are the notions of infinitesimals and they depend on context.

If you just say "an infinitesimal is a real number that is not zero but that is smaller than any number" well of course that isn't a valid definition, as there are no such numbers; but there are various formalisms that allow for things that look like that.

One of them is the formalism of nonstandard analysis, in which for a differentiable function $f$, a real $x$ and an infinitely small $\epsilon$, you have $f(x+\epsilon)=f(x)+ \epsilon f'(x) + \epsilon \delta$ where $\delta$ is also infinitely small; so here the answer is "almost but not exactly". If you want to get rid of the $\delta$ you have to write something like $\frac{f(x+\epsilon)-f(x)}{\epsilon} \approx f'(x)$.

You also have the framework of synthetic differential geometry in which for a function $f$, a "real" $x$ and a $d$ such that $d^2=0$ (these are the infinitesimals of this framework - in this framework, this does not imply $d=0$ ! but you have to reason in a nonclassical logic, because you have to give up on the excluded middle), you have $f(x+d) = f(x) + f'(x) d$. So in this framework, the answer is "kind of yes, if you're willing to change $df$".

Without infinitesimals, it becomes harder to make sense of the question. Indeed, the context where $df$ is defined doesn't allow for things such as $x+dx$ : this is just differential geometry/calculus, and $df$ is just the differential of $f$. In differential calculs you have this (by definition) : $f(x+h) = f(x) + df_x(h) + o(h)$

Of course, the idea of using infinitesimals and writing things such as $f(x+dx) = f(x) + f'(x)dx$ (or $df$), although tricky to make rigourous, is a good one to get some rough idea of what things are and to have some intuition regarding some properties. It can even help in finding proofs of various results (although, if you don't explicitly choose a framework for infinitesimals and use it properly, it cannot be a proof)

Maxime Ramzi
  • 43,598
  • 3
  • 29
  • 104
0

Yes. It is true. Note that $${f(x+dx)-f(x)\over dx}=f'(x)={df\over dx}$$by multiplying the both sides in $dx\ne 0$ we obtain what we want.

Mostafa Ayaz
  • 31,924