Timeline for Summing infinitely many infinitesimally small variables makes sense in algebra
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18 events
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| Sep 28, 2020 at 16:25 | comment | added | Tom Copeland | See "Dual Numbers and Operational Umbral Methods" by Nicolas Behr, Giuseppe Dattoli, Ambra Lattanzi, Silvia Licciardi arxiv.org/abs/1905.09931 | |
| Jun 24, 2020 at 13:16 | comment | added | Anton Mellit | @WillSawin About characteristic $p$ things are a little bit more subtle here. There is this notion of the crystalline site, which uses divided power structures, which when I tried to understand I found unmotivated. I am hoping to find a "physical motivation" through this approach. Maybe something similar is going on here arxiv.org/abs/1803.00812 | |
| Jun 24, 2020 at 13:09 | comment | added | Anton Mellit | @WillSawin From a naive point of view, I like this approach maybe just for aesthetic reasons. When I studied physics, everyone was using things like $f(x+\Delta x)$, and the integral was a sum of deltas. But in analysis it was emphasized a lot that these sort of things "are not allowed", they are not rigorous. So I find it surprising that you can go quite far by using the naive approach of physicists (maybe, just being more explicit about $(\Delta x)^2=0$), if you restrict yourself to pure algebra. For a non-commutative application see my comment about Zassenhaus formula above. | |
| Jun 24, 2020 at 12:38 | comment | added | Will Sawin | @AntonMellit I was trying to argue that your identity is not much different from the usual identity $e^{a+b} =e^{a}e^b$ of formal power series. The characteristic $p$ idea is good, but note that the polynomial algebra doesn't embed into your algebra in characteristic $p$, precisely because $x^n =n!\sum_{i_1< \dots < i_n} \epsilon_{i_1}\dots \epsilon_{i_n}$. | |
| Jun 24, 2020 at 8:06 | comment | added | Anton Mellit | @WillSawin for instance, here is a proof of the Taylor expansion for rational functions. I only use $f(z+\varepsilon)=f(z)+f'(z)\varepsilon$ (for $\varepsilon^2=0$), which is actually the definition. Applying iteratively to $f(z+x)=f(z+\varepsilon_1+\varepsilon_2+\cdots)$ we obtain the following expansion: $f(z+x)=\sum_{n=0}^\infty f^{(n)}(z) \sum_{i_1<\cdots<i_n} \varepsilon_{i_1}\cdots \varepsilon_{i_1}$. In view of $x^n=n! \sum_{i_1<\cdots<i_n} \varepsilon_{i_1}\cdots \varepsilon_{i_1}$ this is Taylor expansion. I didn't have to divide in the proof, which may be useful in char=p. | |
| Jun 24, 2020 at 7:50 | comment | added | Anton Mellit | @WillSawin My question wasn't about proving the identity, it was about its uses. In any case, if you define the exponential using $\prod_i (1+\varepsilon_i)$, then the statement $e^{a+b}=e^a e^b$ is evident. About your second remark: the point to have infinitely many variables is to embed the polynomial ring: $\mathrm{Spec}(R)$ is dense in $\mathrm{Spec}(\mathbb{Q}[x])$ and $\mathrm{Spec}(\mathbb{Q}[[x]])$. For instance, it is a useful principle in comm. algebra: if you need to prove an identity between rational functions it is enough to prove identity of formal power series. | |
| Jun 24, 2020 at 0:18 | comment | added | Will Sawin | This identity is not a perfect analogue of the calculus identity because it doesn't reflect that there are infinitely many variables, just that the variables are infinitely small. The same identity holds for any finite number of variables that square to zero. | |
| Jun 24, 2020 at 0:16 | comment | added | Will Sawin | One has an identity of formal power series $e^{a+b} = e^a e^b$ so $e^{ \sum_{i} \epsilon_i} = \prod_i e^{ \epsilon_i} $ in the ring of formal power series in infinitely many variables, but $e^{ \epsilon_i}=1+\epsilon_i$ if $\epsilon_i^2=0$. | |
| Jun 22, 2020 at 22:05 | comment | added | Anton Mellit | @TomCopeland The analogue of the Trotter formula is $\prod_i (1+(A+B)\varepsilon_i) = \prod_i (1+ A \varepsilon_i)(1+ B \varepsilon_i)$ for some elements $A$,$B$ of a Lie algebra, which in this setting holds trivially because $\varepsilon_i^2=0$. Note that the products are not commutative. It is natural to ask for a proof of Baker-Campbell-Hausdorff, but in this setting another formula, Zassenhaus formula (en.wikipedia.org/wiki/…) follows more easily, by reordering the product. | |
| Jun 22, 2020 at 18:05 | comment | added | Anton Mellit | @BenWieland I don't understand your comment. Did you expect $e^x=\lim_{n\to\infty} (1+x/n)^n$ to hold in $\mathbb{Q}[[x]]$? Of course not, as you explain the limit doesn't even exist. This limit exists in the usual topology on $\mathbb{R}[[x]]$ or $\mathbb{R}$ for fixed $x$. The point of the first identity, because it is well known, is a motivation for the rest. Are you asking if there is a direct connection between this formula and the formula with epsilons? | |
| Jun 22, 2020 at 17:24 | comment | added | Ben Wieland | You start with an identity and leave it hanging. You could delete it without changing the title or rest of post. But since you put it first, it sounds like you want it addressed. You could make sense of that identity in $\mathbb Q[\![x]\!]$. It is false there, because the coefficient of $x^2$ is $(n-1)/2n$, which does not converge in $\mathbb Q$. As @EBz says, $\mathbb Q[\![x]\!]$ embeds in your ring, so it does not change the notion of convergence, so it is still false. | |
| Jun 21, 2020 at 14:47 | comment | added | Tom Copeland | Related to the Trotter product formula (or Lie-Trotter-Kato) of use to physicists, I believe. | |
| Jun 20, 2020 at 17:56 | history | became hot network question | |||
| Jun 20, 2020 at 17:10 | answer | added | Ira Gessel | timeline score: 20 | |
| Jun 20, 2020 at 14:24 | comment | added | user108998 | Just a trivial remark, ur x is a topological nilpotent element of R and so determines an embedding of power series into R, not just polynomials. | |
| Jun 20, 2020 at 11:31 | history | edited | Dan Petersen | CC BY-SA 4.0 |
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| Jun 20, 2020 at 11:13 | history | edited | YCor |
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| Jun 20, 2020 at 9:55 | history | asked | Anton Mellit | CC BY-SA 4.0 |