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May 5[edit]

Do there exist any practical applications of quadruple integrals?[edit]

LaTeX probably includes \iiiint for a good reason. 4-dimensional spacetime would be an obvious guess but I'm not sure that it is equivalent to R⁴ in the usual sense of defining a multiple integral. I Googled and found mention of using quadruple integrals for "determining the interference between elements of a compound array of radiating elements" ([1]), but besides some other obvious cases like four-dimensional Euclidean space, I couldn't really find real-life applications of a quadruple Riemann integral. I am aware that quantum mechanics texts often will use "infinite-dimensional" functional integrals (a concept I currently do not grasp).--Jasper Deng (talk) 09:37, 5 May 2014 (UTC)[reply]

The average distance between two randomly chosen points in the unit square is given as a quadruple integral. Generalisations of this e.g. the weighted average between two random points in some arbitrary shaped area, will have useful applications. E.g. if you have particles that can move freely on a surface and you can treat the interaction in perturbation theory, you can compute the average interaction energy to first order in perturbation theory this way. Count Iblis (talk) 13:22, 5 May 2014 (UTC)[reply]

All (AFAIK) relativistic physical theories can be cast into a stationary action formulation. This action is, for a field theory, a spacetime integral. In smooth manifold theory, the suitable objects to integrate (in a coordinate-independent way) are (often top-dimensional) differential forms. In coordinates, these integrals become (are defined as) ordinary multiple integrals in

R n

. YohanN7 (talk) 13:39, 5 May 2014 (UTC)[reply]

See Lagrangian density for an integral over 4-dimensional spacetime (though the article uses only a single integral sign with d⁴x) --catslash (talk) 22:41, 6 May 2014 (UTC)[reply]

In a machine learning context you'll often want to compute integrals over an arbitrary number of dimensions (e.g., deducing root distribution from leaf observations in a PGM). "Want to" is the key word here since this is often impractical, so other methods are sought. -- Meni Rosenfeld (talk) 12:47, 11 May 2014 (UTC)[reply]

Negative number to the infinity power[edit]

How do we define $n^{\infty }$ and $n^{(-\infty )}$ for negative $n$ ??

For values $n>0$ we have:

$n^{\infty }=\infty$ for $n>1$
$1^{\infty }$ is undefined
$n^{\infty }=0$ for $0<n<1$
$n^{(-\infty )}=0$ for $n>1$
$1^{(-\infty )}$ is undefined
$n^{(-\infty )}=\infty$ for $0<n<1$

But because of how powers of negative real numbers work, defining their infinity and negative infinity powers is much harder. Any thoughts?? Georgia guy (talk) 18:43, 5 May 2014 (UTC)[reply]

I fixed some of your typesetting.

In general, we don't. It doesn't come up enough to matter. If you're doing some work in a particular area where there's a natural choice, give it that definition for the scope of the work. If parts of that work are interesting or applicable to other areas, maybe it'll catch on and we'll start defining things that way.--80.109.80.78 (talk) 22:54, 5 May 2014 (UTC)[reply]

For integer values of n,

n^{\infty }

diverges as n goes to

{(-\infty )}

, so I think it's a fair bet the limit is undefined. AlexTiefling (talk) 23:09, 5 May 2014 (UTC)[reply]

It really depends on whether you regard n as an integer or a real whether 1^∞ is considered as 1 or undefined. Dmcq (talk) 23:39, 5 May 2014 (UTC)[reply]

As suggested by Dmcq, finer points must be brought into play: it is also worth considering whether the exponentiation is considered to be repeated multiplication/division (thus restricting the exponent to integers), or may vary continuously, usually requiring definition through the exponential function. In the latter case, the exponentiation must generally be regarded as undefined for negative bases, and limits even more so. In the former case (restriction to integer exponents), the last bullet should not be considered to hold: the expression diverges rather than converges on

\infty

, for the reason that the expression alternates in sign as the exponent is increases. Essentially, limiting forms such as these depend upon too unstated material (i.e. context) and it makes no sense to attempt a general definition, but given sufficient constraints, a limit can be determined uncontroversially (including possible divergence). —Quondum 23:47, 5 May 2014 (UTC)[reply]

Yes that's a better way of going about it. Is it repeated multiplication or a use of the exponential function. Dmcq (talk) 11:48, 7 May 2014 (UTC)[reply]