Is it possible to generalize functions like $x^y, \ln x, \sin x, \arctan x$ to surreal numbers or surcomplex numbers? Which of their properties and relations (e.g. usual trig identities) will still hold in this case? Is it possible to perform differentiation and integration on these generalized functions?
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$\begingroup$ The set theory tag would be better as a logic tag. This is really the area of nonstandard analysis. $\endgroup$– Jason RuteCommented May 25, 2013 at 5:07
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$\begingroup$ I will also comment that this is also the area of model theory over metric structures. You should be able to get hyperfinite nonstandard models of the real and complex numbers with these functions by taking ultraproducts. What I can't remember is (1) which properties are preserved by ultraproducts of metric structures, and (2) how exactly ultraproducts of $\mathbb{R}$ and $\mathbb{C}$ relate to surreal numbers. For example, is the ultraproduct $\prod_I \mathbb{R}$, where I is a proper class, isomorphic to the surcomplex numbers? I'll leave these questions to the experts on this subject. $\endgroup$– Jason RuteCommented May 25, 2013 at 5:23
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$\begingroup$ For trig functions, see mathoverflow.net/questions/91337 $\endgroup$– Gerald EdgarCommented May 30, 2013 at 19:09
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2 Answers
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The following is taken from a sci.math post by Nicolau C Saldanha. Much more detail is available there. It is important to note that Nicolau does not define functions extensionally, i.e. two functions which agree everywhere but are defined differently are considered distinct.
We use the convention $f(x) = \{ f_L(x,x_L,x_R) | f_R(x,x_L,x_R) \}$ for surreal functions (note that this already breaks extensionality). We can define the definite integral of a surreal function as follows:
$$\begin{align} \int_a^b f(t) dt = \left\{\int_a^{b_L} f(t) dt + \oint_{b_L}^b {f_L}(t) dt,\; \int_a^{b_R} f(t) dt + \oint_{b_R}^b {f_R}(t) dt,\right.\\\\ \int_{a_R}^b f(t) dt + \oint_a^{a_R} {f_L}(t) dt,\; \int_{a_L}^b f(t) dt + \oint_a^{a_L} {f_R}(t) dt\huge\mid\\\\ \int_a^{b_L} f(t) dt + \oint_{b_L}^b {f_R}(t) dt,\; \int_a^{b_R} f(t) dt + \oint_{b_R}^b {f_L}(t) dt,\\\\ \left.\int_{a_R}^b f(t) dt + \oint_a^{a_R} {f_R}(t) dt,\; \int_{a_L}^b f(t) dt + \oint_a^{a_L} {f_L}(t) dt\right\} \end{align}$$
Note: $\oint$ denotes direct integration, meaning the integral is evaluated without chopping the domain into pieces, not countour integration. Properly the integral should have a capital $D$ superimposed on it, but I am not sure how to do this on this site.
The definition of integral gives you $\ln x$ right away: $\ln x = \int_1^x \frac1t dt$. The integral also lets you solve the differential equations $y'=y$ and $y''=-y$, giving you $e^x,\sin x$ and $\cos x$. Nicolau claims that most standard functions, and even functions like the Riemann Zeta function should be obtainable in this manner.
However, this definition has its problems, which are discussed briefly in the linked sci.math thread. In particular, it does not behave nicely with the surreal exponential function.
An exposition on the surreal exponential can be found in Harry Gonshor's "An introduction to the Theory of Surreal Numbers".
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14$\begingroup$ What on earth is this direct integration supposed to be? All you've said about it is that I should not chop the domain into pieces, whatever that means... $\endgroup$– zebCommented May 25, 2013 at 18:11
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$\begingroup$ Trial and error for the spacing:
\int\hspace{-1.8ex}Dproduces $\int\hspace{-1.8ex}D$, and\int\hspace{-2.2ex}Dproduces $$\int\hspace{-2.2ex}D$$ (at least in the answer box; we'll see how it works in comments). $\endgroup$– LSpiceCommented Jul 28, 2017 at 0:45
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(Too long for a comment.)
It would seem to me that one other possibility for defining a transcendental function over the surcomplexes might be to use a suitable modification of the Cauchy integral formula; since you can reciprocate surcomplex numbers, one could then consider
$$f(z)=\frac1{2\pi i}\oint_\gamma f(t) (t-z)^{-1}\mathrm dt$$
where $\gamma$ is some suitable anticlockwise contour. I understand that this is on the surface a naïve proposal, and I would be interested in hearing how this might break down.