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Let $D$ be the ring $\mathbb{C}[\epsilon]/\langle \epsilon^2\rangle$. Define the functions $dual : \mathbb{C} \to D$ and $stdPart : D \to \mathbb{C}$ by $dual(x) := x+0\cdot \epsilon$ and $stdPart(x+y\cdot \epsilon) := x$, respectively. Define $exp : D \to D$ and $ln := D-\langle \epsilon \rangle \to D$ by $exp(x+y\cdot \epsilon) := \exp(x)+\exp(x)\cdot y\cdot \epsilon$ and $ln(x+y\cdot \epsilon) := \ln(x)+\frac1x \cdot y \cdot \epsilon$, using the principal branch of $\ln$.


Let $f$ be an expression using addition, negative, multiplication, division, the variable $z$, constants from $\mathbb{C}$, $\exp$, and $\ln$. Let $g$ be the result of starting with $f$ and replacing $z$ with $0+1\cdot \epsilon$, $\exp$ with $exp$, $\ln$ with $ln$, and each constant $c$ with $dual(c)$. Suppose we evaluate $g$ such that we

(a) don't $ln$ any member of $\{d\in D : \operatorname{Re}(stdPart(d))\leq 0$ and $\operatorname{Im}(stdPart(d)) = 0\}$
(b) don't divide any member of $D$ by $0+0\cdot \epsilon$
(c) don't divide any member of $D-\langle \epsilon \rangle$ by any member of $\langle \epsilon \rangle$
(d) do evaluate $\frac{0+x\cdot \epsilon}{0+y\cdot \epsilon}$ as $\frac{x}{y}+0\cdot \epsilon$
(e) don't do nested divisions of members of $\langle \epsilon \rangle$ by members of $\langle \epsilon \rangle$
(f) do get the result $L$


Does it follow that, using the principal branch of $\ln$, $\; \; \displaystyle\lim_{z\to 0} \: f = stdPart(L) \,$ ?

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You may enjoy googling around for "synthetic differential geometry". –  Theo Johnson-Freyd Nov 4 '10 at 3:32
    
Another thing to do is re-write in little-o notation... $a+b*epsilon$ becomes $a+bz+o(z)$ where $z \to 0$. –  Gerald Edgar Nov 4 '10 at 4:09
    
There is quite a dearth of findable proofs when googling "synthetic differential geometry", which is why I asked this question in the first place. –  Ricky Demer Nov 4 '10 at 5:43

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