# Learning differential calculus through infinitesimals

In class, we've studied differential calculus and integral calculus through limits. We reconstructed the concepts from scratch beginning by the definition of limits, licit operations, derivatives and then integrals. But the teacher really did everything to avoid talking about infinitessimals. For instance when we talked about variable changes we had to swallow that for a certain $u=\phi(x)$, $du = \phi'(x)dx$. What bothers me is how much we use them in maths and physics without understanding them. My question is how can I learn to manipulate infinitessimals? I've read an article on the field of infinitessimals but it didn't quite satisfy my curiosity. What literature would you recommend on the subject knowing that I have some background on linear algebra (vector spaces, matrices), algebraic structures (groups, rings, fields) and monovariable calculus.

• Prepare to ping pong those diferentials left and right on Physics – Dleep May 22 '16 at 9:43
• We definetly do but i have no idea how to rigorously use them! We unscrupulously divide, multiply and add them together! – Badr Youbi Idrissi May 22 '16 at 9:47
• Just think of $dx$ and $df$ as extremely tiny numbers (but not "infinitely small", because it's unclear what that means without nonstandard analysis). We can call them $\Delta x$ and $\Delta f$ if we like. $\Delta x$ is a tiny change in $x$, and $\Delta f$ is the corresponding change in the value of $f$. (So $\Delta f = f(x + \Delta x) - f(x)$.) Note that $f'(x) \approx \Delta f / \Delta x$ (and the approximation is very good). Equations you derive by manipulating $\Delta x$ and $\Delta f$ are only approximately true, but we can hope that "in the limit" we obtain true equality. – littleO May 22 '16 at 10:57
• Since you mention algebra, look at the dual numbers -- i.e. study the quotient ring $R = \mathbb{R}[\epsilon] / (\epsilon^2)$. e.g. one fact about this ring is that if $f$ is a polynomial over $\mathbb{R}$, then $f(x + \epsilon) = f(x) + \epsilon f'(x)$, where the derivative is the formal derivative: the linear map that sends $x^n \mapsto n x^{n-1}$. $\epsilon$ is called a "nilpotent infinitesimal", since some power of $\epsilon$ is exactly zero. – user14972 May 22 '16 at 13:30
• The keyword "differential form" is a good one to know; it's what $dx$ and $du$ are usually taken to be - and it basically just says they are vector fields on some manifold - and in it, you get that relations like $du=\phi'(x)\,dx$ can actually be true. I find this very satisfying. – Milo Brandt May 22 '16 at 15:03

Hope this relevant to what was asked...

To focus on $u = f(x), du = f'(x) dx,$ take this as an easy way to write $u = f(x), \Delta u = f'(x) \Delta x,$ where $\Delta x$ is an interval in $x$ that we can make arbitrarily small. This formula makes sense for $f$ smooth because in a small enough region around $x,$ $f$ is locally linear.

Recall that you can write $f(x) \approx f(x_0) + (x-x_0) f'(x_0) + {\cal O}((x - x_0)^2).$ Making $\Delta x \equiv x-x_0,$ and $\Delta f = f(x) - f(x_0),$ the original formula is obtained by ignoring the ${\cal O}(\Delta x^2)$. These definitions make sense because we have assumed that $f$ is smooth, so $\Delta f$ varies in proportion to $\Delta x$ (think of the delta-epsilon definition of continuity). None of this requires infinitesimals, it just requires continuity of the function and intervals that we can make "small enough". This is why multiplication by quantities like $dx$ makes sense.

Dealing with "actual" infinitesimals is the field of nonstandard analysis.

• I would write $\Delta u \approx f'(x) \Delta x$ (not $=$), because the sides are not exactly equal except within intervals where $f$ is exactly linear; but $f(x) = f(x_0) + (x-x_0) f'(x_0) + {\cal O}((x - x_0)^2)$ (not $\approx$), because the term ${\cal O}((x - x_0)^2)$ is able to exactly account for the error in the approximation. – David K May 22 '16 at 15:09
• Not sure why these edits were made by @Hurkyl: $f(x)\approx f(x_0)+(x−x_0)f′(x_0)+{\cal O}((x−x(0)^2)$ should be an equality (=), it is not an approximation. Also, ${\cal O}(x^2)$ is equivalent to what the editor wrote: ${\cal O}((x - x_0)^2).$ If anything, ${\cal O}(x^2)$ is more correct, for the same reason one would not write ${\cal O}(\frac{x^2}{2}).$ – Merkh May 23 '16 at 9:54
• @Merkh: $O(x^2) = O((x-x_0)^2)$ as $x \to \infty$, but this example isn't about the asymptotics as $x \to \infty$: It's about the asymptotics as $x \to x_0$. As $x \to x_0$, $O(x^2) = O(1)$. (assuming $x_0 \neq 0$). – user14972 May 23 '16 at 14:08
• Ah yes very correct, my mistake. Good call! – Merkh May 23 '16 at 19:52

The usual treatment of calculus today avoids using infinitesimals entirely. $\mathrm{d}x$ is merely a notation that does not literally refer to an infinitesimal. That said, there is an alternative approach that uses infinitesimals known as nonstandard analysis. The Wikipedia articles on nonstandard analysis and nonstandard calculus are probably good places to start if you just want to find out if the topics interest you. Be aware these systems are not the usual treatment of calculus and people will be confused if you do not explicitly state you are working in them.

• What does dx stand for then if not infinitessimals? – Badr Youbi Idrissi May 22 '16 at 10:52
• The usual thought process is the approach of @littleO in a comment to your question. – GPhys May 22 '16 at 11:07
• @BadrYoubiIdrissi - It's complicated. You will have to read about differential forms. – steven gregory May 22 '16 at 15:42

The best calculus textbook based on infinitesimals is Keisler's textbook Elementary Calculus. This opinion is based on our experience teaching calculus with infinitesimals based on the book over the past three years. We have taught over 250 students by now using this method, yielding better results than parallel groups that did not use infinitesimals, based on exam scores and teacher evaluations at the end of the course.

• I will have a look at the text book, thank you. – Badr Youbi Idrissi May 22 '16 at 12:31