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In one of Numberphile's videos they describe the $-\frac{1}{12}$ as a "gold nugget" inside the sequence $1+2+3+4+\dots$ surrounded by a bunch of "rock" that is infinity.

Can these numbers that we get with regularization/analytic continuation of Reimann zeta/etc be described more simply using the hyperreals? Is there some particular non-arbitrary unlimited hyperreal $A$ you can associate with $1+2+3+4\dots = A-\frac{1}{12}$? Like in a Casimir force calculation problem, where the forces on both sides are infinite but differ by some finite value (related to the $-\frac{1}{12}$), is there some specific, non-arbitrary unlimited hyperreal value $A$ associated with the infinite part of that force?

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closed as off-topic by Andrés E. Caicedo, Xander Henderson, JonMark Perry, José Carlos Santos, user223391 Jun 13 '18 at 0:22

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No. Robinson's hyperreals just are a different way of getting at the same results that you could get with traditional $\varepsilon$-$\delta$ definitions. Since the sum traditionally diverges, that's what hyperreals tell you.

Now, you could probably formalize your favorite way of getting to $-1/12$ with the hyperreals in place of whatever limits you'd calculate, but that would not be a method at getting the value "more simply".

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  • $\begingroup$ I'd be happy to add details/clarify something if the downvoter would care to share what they thought was lacking. If you don't want to comment publicly, you can use the contact in my profile $\endgroup$ – Mark S. Jun 10 '18 at 10:58
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The reason the expression $\sum_{i=0}^\infty a_i$ makes sense, is because if it converges, the point of convergence is unique.

So, for any converging series $(a_i)_{i\in\mathbb{^*N}}$ we also have $\sum_{i=0}^{h_1} a_i \approx \sum_{i=0}^{h_2} a_i $ , for $h_1,h_2\in\mathbb{^*N-N}$. However, if a series is diverging, then $\sum_{i=0}^{h_1} a_i - \sum_{i=0}^{h_2} a_i $ can have infinitely many results depending on your choice of $h_1,h_2$.
So while for every $h\in\mathbb{^*N-N}$ we can find an $A\in \mathbb{^* R}$ so that $\sum_{i=0}^{h} i = A-\frac 1 {12}$ holds, there can be no constant $A$ that fulfills this equation for all possible $h\in\mathbb{^*N-N}$.

While infinite and infinitesimal values in nonstandard analysis exist, their nature is rather analytical. As we can't construct explicit infinite values, their main use is that all finite values from their perspective "look the same".
As example: For any infinite number $h$ and any finite number $n$, the equation $\frac n h \approx 0$ holds.

So, to pick up your example of Casimir-force: Whether we are in standard- or nonstandard-analysis, we don't actually want to discriminate infinite values, we merely want to show and use the fact that these infinite values are greater than any finite value.
So, if we had two infinite sums whose difference is finite, in both standard- and nonstandard-analysis we'd look at the difference of both sums rather than at each sum individually.

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    $\begingroup$ It is not true that "for every $h\in\mathbb{^*N-N}$ we can find an $A\in \mathbb{^* N}$ so that $\sum_{i=0}^{h} i = A-\frac 1 {12}$ holds". This is because by transfer all partial sums are hyperintegers, and therefore cannot differ from a hyperinteger by a fraction like $\frac{1}{12}$. This problem cannot be treated without dealing with analytic continuation. $\endgroup$ – Mikhail Katz Jun 10 '18 at 15:49
  • $\begingroup$ @MikhailKatz Ahh, missed that one, corrected it $\endgroup$ – Sudix Jun 10 '18 at 17:19
  • $\begingroup$ I think it's worth pointing out that defining $\sum_{i=0}^h a_i$, for $h$ nonstandard, is a tricky issue in itself. Further, I believe it's not always well-defined, only when the sequence $a_i$ can be "internally defined", though I don't have background to make that precise. $\endgroup$ – Wojowu Jun 10 '18 at 17:28
  • $\begingroup$ @Wojowu The easiest way to define $\sum_{i=0}^h a_i$ is by making it a shorthand for a partial sum. So, $\sum_{i=0}^n a_i = a_1 + ... + a_n$. Then we make a function out of it, $a: \mathbb{N} \Rightarrow \mathbb{R}$ , $a(k) = \sum_{i=0}^k a_i$. Per transfer we therefore can now put in any hypernatural value $h$ and make sense of $\sum_{i=0}^h a_i$ $\endgroup$ – Sudix Jun 10 '18 at 20:21
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    $\begingroup$ Every sequence $a:\mathbb N \to \mathbb R$ has a natural extension ${}^\ast: {}^\ast\mathbb N \to {}^\ast\mathbb R$ and in particular $a_H$ is defined for each infinite hyperinteger $H$. Similar remarks apply to partial sums. Thus infinite partial sums are defined for every series one starts with. $\endgroup$ – Mikhail Katz Jun 12 '18 at 8:54

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