Does $f(a+b\epsilon)=f(a)+b\epsilon f'(a)$ remain true for non-analytic smooth functions of dual number argument?
Where $ \epsilon^2=0$, $\epsilon \neq 0$ and $a,b \in \mathbb R$. I found proofs based only on Taylor series. That means it is okay to use the formula for argument values where function is analytic (in real numbers). But what about other cases?

1. Does a proof of $f(a+b\epsilon)=f(a)+b\epsilon f'(a)$ for non-analytic smooth functions exist?

2. What about dual values $b\epsilon g(a)$ that correspond to argument $a$ where function is non-analytic?

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  • $\begingroup$ This sounds really cool -- do you have a reference for where one can learn about non-standard complex analysis like the above? (hypercomplex numbers) I only know of textbooks about non-standard analysis for the reals. $\endgroup$ – Chill2Macht Jun 18 '16 at 12:52
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    $\begingroup$ Sorry I meant $a$ and $b$ are real numbers. I thought that use of hypercomplex numbers in analysis is non-standard analysis. I am correcting that. $\endgroup$ – Alex Alex Jun 18 '16 at 13:00

For an analytic function of a real input, it's very natural to use the power series to define it on arbitrary dual numbers, and then your result about "automatic differentiation" falls out immediately.

But for a function that's not analytic, it's not generally obvious how to define the function on nonreal inputs, and whether or not you have a similar result depends on how you choose to extend the smooth function to the dual numbers.

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  • $\begingroup$ I am interested in smooth non-analytic $\mathbb R \to \mathbb R$ functions constructed from analytic ones by addition, subtraction, multiplication, division and composition. $\endgroup$ – Alex Alex Jun 18 '16 at 16:03

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