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[Virex]

Theoretical Math Question!

The following formula (the surveyor's formula) can be used to calculate the area of any polygon:
n-1
 Σ  (x_iy_i+1 - x_i+1y_i)/2
i=0

Does a generalization of this formula to the third or higher dimensions exist? If so, what is it?

for 3 dimensional figures, yes. There ought to be a generalization to more dimensions, but to find that you'll first need to extend the divergence theorem to more then 3 dimensions.

[Vector]

Already done by Stokes, IIRC.  The so-called "Stokes' Theorem" is actually far more generalized than they usually teach it in Calc III.

(Don't take my word for it--this was an aside in a course I took 3 years ago)

[Christes]

Mathematical object.

I figured it might be good for a laugh.

We actually talked about this in algebraic topology (together with some less well-behaved variants) as an intro to homotopy.  It's funny how you get used to the strange names after a while.  (I don't even snicker at the hairy ball theorem anymore.)

Already done by Stokes, IIRC.  The so-called "Stokes' Theorem" is actually far more generalized than they usually teach it in Calc III.

(Don't take my word for it--this was an aside in a course I took 3 years ago)

That's certainly possible.  I've only seen the (general) Stoke's Theorem done in terms of abstract differential forms, with unfortunately little appeal to geometry.

[squeakyReaper]

I'm looking for some sort of app, site, or program that will take data and give me an approximate fit for a logarithmic equation.  I have the numbers I want to use, and I know they don't fit exactly on a line.  Is there anything you guys know that could work?

Edit;  Found this, so I guess this post can be ignored.

Double Edit:  It doesn't work.  Back to square one.  You guys know anything?  I see something saying there's a graphing calculator that can do it, but that's a phsyical one, not something I can use right now...

[Virex]

I can't think of a free program of the top of my head, but I just came upon an on-line version of the statistics tool I often use for this kind of work. After loading your data (pray you're under the limit of 100 elements because it's still a preview version), the option you're looking for is under relate-simple regression. Then you ought to be able to find an option to switch to log(x)-y or something similar. It'll also tell you how good the fit is.


Edit: you'll find it under analysis options. It's called logarithmic-x and is the second option of the middle row.

[Argembarger]

Gaussian integral

what is this I don't even

so beautiful

[Vector]

Isn't it astonishing and beautiful?

[Darvi]

Funny how limits to infinity can be so simple. Well, if you consider irrational numbers simple.

[ZetaX]

Well, is e+pi irrational¿

[Darvi]

I finally understand why e^(ix) = cos x + i sin x

Edit:

Well, is e+pi irrational¿

Yes it is.

[Christes]

Err, I'm not so sure about that.  Wolfram is just mimicking what you wrote.  One can easily prove, for instance, that one of either e+pi or e-pi must be irrational.  But proving that one of them actually is irrational probably hasn't been done.  I'd love to see a proof if it has, however.

[Vector]

... We're just proving that both e + pi and e - pi are irrational?

If we assume that both e and pi are irrational, then that's exercise 1 in Baby Rudin, my friend.

[Christes]

Exercise 1 in Baby Rudin is when one of the two is rational.  (The sum of two irrational numbers is not, in general, irrational: take sqrt(2) and -sqrt(2) )

[Vector]

Huh, wow.  I'm a moron.

Thanks for the reminder =/  Now I really have no idea how that'd be proven.

[Christes]

I seriously think it might still be open.  A lot of these types of problems are.  I mean it just has to be irrational, right?

But actually getting your hands around things like e+pi and e*pi is really hard.  The proofs for e and pi use integration tricks and/or formulas that are very specific to the constants e and pi.  There probably aren't too many profound formulas for e+pi.

Note: e^pi apparently is transcendental by this theorem.  Awesome - I hadn't realized that.