Wikipedia:Reference desk/Archives/Mathematics/2011 December 27

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December 27

in what way is the sum of all positive integers -1/12

per hardy and ramajamanan. I don't know much math but it's pretty obvoius to me that no amount of positive integers can add up to a negative number. — Preceding unsigned comment added by 80.98.112.4 (talk) 01:45, 27 December 2011 (UTC)

Well, it depends on what sort of addition you have in mind. We have an article on 1 + 2 + 3 + 4 + …. You're certainly correct that, in the usual sense in which infinite sums are interpreted in mathematics, this sum is infinite. --Trovatore (talk) 01:58, 27 December 2011 (UTC)

Right that article doesn't specify the special way/approach under which it isn't. I do understand infinities can be weird, for example in one partcular approach there are as many positive integers as positive and negative integers (1:1, 2:-1, 3:2, 4:-2 and and infinitum) but in another way there is one more positive integer than positive and negative integers: (1: no correspondence, 2:1, 3:-1, 4:2, 5:-2 etc.) it dependa on how you count.

so how can you count sch that you converge on 1/12? 188.156.156.39 (talk) 02:10, 27 December 2011 (UTC)

sorry, not 1/12 (weird enough, since they're all whole numbers and no operation like division is involved, just adding, but it's not even 1/12, but -1/12. how? (under what approach). 188.156.156.39 (talk) 02:14, 27 December 2011 (UTC)

You can't, in the ordinary sense. Any finite subset of the terms has positive sum, and in any usually applied topology, those sums either do not converge on anything, or converge to +∞

However, there is a function called the Riemann zeta function, that in certain parts of the complex plane equals the following sum:

$\zeta (s)=1^{-s}+2^{-s}+3^{-s}+\ldots$

Now, if you substitute −1 in for s, of course the series diverges. However, if you take the function where the series does converge, and travel around the complex plane to get to −1, avoiding poles, you get to the value

$\zeta (-1)={\frac {-1}{12}}$

Now, is this enough to say that the sum is −1/12? In my opinion, no, usually. But there are a few special circumstances where it seems to make sense. --Trovatore (talk) 02:22, 27 December 2011 (UTC)

(edit conflict) By Zeta function regularization. The idea is that the sum $\sum _{n=1}^{\infty }n^{-s}$ defines an analytic function of s. This can be analytically continued to points outside the domain of where the summation makes sense. In particular, the value of its analytic continuation at $s=-1$ is $-1/12$. It's not, however, the same thing as the infinite series $\sum _{n=1}^{\infty }n$, since this properly diverges to infinity. Sławomir Biały (talk) 02:23, 27 December 2011 (UTC)

The thing that really needs to be explained is, why should anyone interpret the sum as an instance of $\sum _{n=1}^{\infty }n^{-s}$ at s=−1, rather than, say, $\sum _{n=1}^{\infty }n^{-{\frac {s}{2}}}{\frac {s^{n}}{(-2)^{n}}}$ at s=−2? I haven't checked but I assume that gives you a different answer. Seems very ad hoc, not remotely natural. The surprising thing is that there are a few contexts, not obviously related, where it seems to be the right answer. --Trovatore (talk) 02:32, 27 December 2011 (UTC)

There's some sense it which the zeta function is the natural regularization from the point of view of Fourier analysis and spectral theory. The zeta function of an elliptic operator A is the trace of the kernel of A^s. (Here A^s is usually already defined by analytic continuation. The trace is regarded in a suitable distribution sense which effectively means that the trace operation commutes with analytic continuation.) The zeta function is natural because it contains most of the asymptotic information of the spectrum (think the prime number theorem). For instance, the Atiyah-Singer index is expressible in terms of zeta functions, and the Wiener–Ikehara theorem describes the asymptotics of the eigenvalues. Sławomir Biały (talk) 13:02, 27 December 2011 (UTC)

What is it called when an irrelevant outcome between 3rd place and 4th place determines who finishes 1st or 2nd?

I took a creative mathematics class back in high school that discussed how in ice skating, that depending on how well (or how badly) the last contestant performs--even if that candidate has no possibility of finishing first or second place--can flip the ranking of other ice skaters. I looked for it on the article Apportionment paradox but I don't know what it is called, or where to begin looking. It's when an athlete indirectly influences the ranking system in bizarre ways, such as if Alex is in first place, Brad is in second place, Charlie is in third place, and if Donald moves from 4th place to 3rd place, then suddenly Brad has more points then Alex. (I think it has to do with competitions which have a running total, and points are awarded based by relative ranking in the component events) I found it fascinating because it is something similar to what's actually happening in a company I do business with regarding how it awards contracting, so any guidance would be tremendously appreciated! Thank you all, 완젬스 (talk) 06:24, 27 December 2011 (UTC)

See Independence of irrelevant alternatives. Additionally, see Arrow's impossibility theorem. 77.125.138.22 (talk) 10:45, 27 December 2011 (UTC)

Yes! Thank you very much, that's it. 완젬스 (talk) 12:31, 27 December 2011 (UTC)