Let $f : \mathbb R \to \mathbb R$, and consider the differential equation $$ f'(t) = f(t) $$ it is easily seen that it has the solutions $f(t) = a\cdot \exp(t)$ for $a \in \mathbb R$. Now another way of writing this $$ \frac{df}{dt} = f $$ (here the argument isn't written anymore, an abstraction step leading to the next notation). Seeing $\frac{df}{dt}$ as a shorthand for $f'(t)$ this is no problem, but I have seen the notation $$ \frac{df}{f} = dt. $$ Suddenly we have two new quantities and $df/dt$ gets handled like a fraction (not just merely a notation for $f'$ anymore), I know this is the Leibniz way of thinking about differentials, and I knew it for example as a shorthand from the substitution rule for integrals.

I know how to think about such terms, $dt$ should be interpreted as $t_{i+1} - t_i$ for two time steps or $(t + h) - t$ as the time steps get finer, or $h \to 0$, so $dt \approx h$, similarly $df$ could be understood as $f(t + h) - f(t)$ in the limit, so $df \approx f(t+h) - f(t)$ for $h$ sufficiently small at $t$.

But how should $df / f = dt$ be read, as "the proportion of the change in quantity to the total quantity equals the change in time, i.e. for constant time changes it is constant too, which implies that if the quantity increases, the rate of change has to increase too." But with substituting $$ df \approx (\exp(t + h) - \exp(t)) \mbox{ and } dt \approx (t + h) - t $$ we should have $$ \frac{\exp(t + h) - \exp(t)}{\exp(t+h)} = h $$ as $\exp(t+h) - \exp(t) \approx \exp(t+h)h$ (i.e. $df = fdt$) which is not valid, because $(\exp(t + h) - \exp(t))/\exp(t+h) = 1 - \exp(-h) \ne h$? So where is the problem, why does this interpretation does not work?

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    $\begingroup$ Well, but $1 - \exp (-h) = h + O(h^2)$, so up to first order you have equality. However, can you read $\frac{df}{f} = dt$ as an equality of differential forms? If you can, you have a nice framework where everything works out fine and rigorously. $$\frac{df}{dt}(t) = f(t) \leadsto \frac{df}{dt}(t)\,dt = f(t)\,dt \leadsto (df)(t) = f(t)\,dt \leadsto \frac{df}{f}(t) = dt.$$ $\endgroup$ – Daniel Fischer Jul 8 '15 at 11:28
  • $\begingroup$ Okay, these "differential equations and rearrangements" are just meant "up to first order terms". No, I have no knowledge about differential forms, if it is not to complicated you might expand your comment to an answer, maybe say something about differential forms, so I can check this thread as answered... $\endgroup$ – StefanH Jul 8 '15 at 11:39
  • $\begingroup$ No, they are exact, if properly interpreted. But if you start interpreting them as $df \approx \exp(t+h) - \exp (t)$ and so on, you either use it as a heuristic tool, then you can ignore everything above some order [which need not always be the first order], or you make a rigorous theory of infinitesimals where these manipulations are legitimate and exact. In my opinion, infinitesimals are too unintuitive to be worth the hassle, but others find them intuitive. Well, or you put a $\lim\limits_{h\to 0}$ everywhere, and you are back in standard analysis land. $\endgroup$ – Daniel Fischer Jul 8 '15 at 11:54
  • $\begingroup$ You mean instead of $dt \approx \exp(t+h) - \exp(t)$ in standard analysis land we would have $df = \lim_{h\to 0} (\exp(t+h) - \exp(t)) = 0$, would not make any sense... $\endgroup$ – StefanH Jul 8 '15 at 12:07
  • $\begingroup$ No, we then need to divide everything by $h$ to get something useful. Of course that lands us square where we started here, $\frac{f'(t)}{f(t)} \equiv 1$. $\endgroup$ – Daniel Fischer Jul 8 '15 at 12:10

Everything works out when you use integrals instead. For example, start with $$ \int \frac{f'(t)}{f(t)}dt = \int dt. $$ Now, use a change of variables on the left of $y=f(t)$, $dy = f'(t)dt$: $$ \int \frac{dy}{y} = \int dt. $$ So, whether you play the same kind of game with a change of variable, or you do the manipulation directly on the equation, it works out the same way. So you learn a formalism that saves time, but has a justification underlying it. It takes some time to convince yourself that formalism works for integrals, too, but it does. It's good that you're thinking about it, though, because there is always a potential to use symbolic manipulation where it does not apply.


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