How do you work with infinitesimal exponents in synthetic differential geometry?

Question

I just read this paper by Andrej Bauer, which discusses the basic tenets of synthetic differential geometry. Namely, that for any function $f$, any real number $x$, and any infinitesimal $\epsilon$ (a number which squares to $0$), there is a unique number $f'(x)$ such that $$f(x + \epsilon) = f(x) + f'(x)\epsilon$$

This allows us to simply prove some of the most useful high-school deriviatives. But, I am wondering about differentiating the function $f(x) = e^x$. Plugging it in, I get $f(x + \epsilon) = e^{x + \epsilon}$, but I'm not sure how to reduce this into the form $e^x + e^x\epsilon$. In general, how do we handle infinitesimal exponents $a^\epsilon$?

Answer 1

Andrej Bauer has answered this (good!) question before on his blog. I will just quote him (and make this post community wiki):

Before you write down an infinite sum, like you did [resp. like it was done in the other answers], you have to argue that they make sense. Which they don't in synthetic differential geometry. There are no axioms that ensure completeness of the smooth real line with respect to Cauchy sequences. Nor can there be such axioms, because they would allow us to define a non-smooth function as an infinite sum.

Actually, the exponential function is introduced as the solution of the differential equation $y′=y$ with the initial condition $y(0)=1$, so the derivative of $e^x$ is postulated to be $e^x$.

Answer 2

I don't know synthetic differential geometry, so maybe this is just a guess: $$ e^{x+\varepsilon} = e^x e^\varepsilon = e^x\left( 1+\varepsilon+\frac{\varepsilon^2}{2}+\frac{\varepsilon^3}{6}+\frac{\varepsilon^4}{24} +\cdots\right) $$

Then use the fact that $\varepsilon$ squares to $0$.

Answer 3

Expand as a power series in $\epsilon$ around the point $0$.
$$\mathrm{e}^{x+\epsilon} = \mathrm{e}^x + \mathrm{e}^x \epsilon + \epsilon^2\left( \frac{\mathrm{e}^x}{2} + \epsilon \frac{\mathrm{e}^x}{6} + \dots \right) = \mathrm{e}^x + \mathrm{e}^x \epsilon$$

Likewise, $$a^{x+\epsilon} = a^x + a^x \ln(a) \epsilon + \epsilon^2\left( \frac{a^x \ln a}{2} + \epsilon \frac{a^x \ln a}{6} + \dots \right) = a^x + a^x \ln(a) \epsilon$$

Although, once you have the former, it's easier to use it and $a^x = \mathrm{e}^{x \ln a}$ than to just churn out the latter.

Answer 4

I'm just learning synthetic differential geometry, so forgive me if this isn't what you're looking for.

We may rewrite $a^\epsilon=e^{\epsilon\ln a}$. Thus, we have, $f(x):=e^{x}$: $$e^{x+\epsilon}=e^x+f^\prime_{\mathrm{synthetic}}(x)\epsilon\\ \implies \dfrac{e^x\left(e^{\epsilon}-1\right)}{\epsilon}=f^\prime_{\mathrm{synthetic}}(x)$$ As $\epsilon\to0$, we have $$\lim_{\epsilon\to0}\dfrac{e^x\left(e^{\epsilon}-1\right)}{\epsilon}=\lim_{\epsilon\to0}{e^x\left(e^{\epsilon}\right)}=e^x=f^\prime_{\mathrm{synthetic}}(x)$$ This implies that $$e^{x+\epsilon}=e^x+f^\prime_{\mathrm{synthetic}}(x)\epsilon\\ e^{x+\epsilon}=e^x+e^x\epsilon$$