Recent content by PhilG

The way we were playing it, the numbers did not have to come in any particular order. It's okay to roll numbers other than the six place numbers or seven, and also okay to repeat numbers that have already come up. So 4, 5, 6, 8, 9, 2, 10 would count. So would 2, 9, 5, 6, 8, 8, 8, 6, 4, 12, 10...

Thanks for those calculations, mathwonk and gnome. Wow, 14 to 1 huh? I would have thought the odds against throwing the numbers in a specific order would have been lot higher, like 10000 to 1 or something. I just thought the guy was having an amazing streak of luck. Maybe I don't have any...

I was playing dice at a bar tonight, and after a while one of the guys I was playing asks "What do you think are the odds against rolling all the place numbers (4, 5, 6, 8, 9, 10) before rolling a seven?" So I'm thinking, just rolling a 4 before a 7 is a 2-to-1 shot, so rolling all six numbers...

Mine was 103 in 2nd grade. But what I lack in IQ, i make up for in penis length.

A Brief on Tensor Analysis by James Simmonds is very good.

First things first... See if your library has a copy of A Brief on Tensor Analysis by James Simmonds. The author's background is in continuum mechanics, if I recall correctly. Here's a start on the upper and lower indices stuff, following Simmonds' approach. In three dimensional euclidean...

good for you, buddy.

Yes, that's right. But you have to prove that A is equal to the union of those U's.

By definition, X and the empty set are open in X. Also, an arbitrary union of open sets is open, and a finite intersection of open sets is open.

Is that last term "the vector h, scaled by a number that goes to zero faster than |h|"? What if h and f(h) live in spaces of different dimensions, say n and m?

yokebutt: I tend to agree with Kino that determining the pressure experimentally is probably the best way to go. However, it is also a good idea to have a simplified mathematical model that gives you a ballpark answer. Here is a formula that I think might work: Pressure = (Total...

Limits are defined for general topological spaces, even ones that don't have a metric. I'm not sure how to define a derivative without a norm though. Suppose you have a function f:Rn-->Rm It seems as though you need a way to associate a number with each displacement vector h in Rn, i.e. a...

matt: So the answer to my question is yes?

f(x0) is not necessarily in V. Also, is a topological vector space really all the structure you need to do calculus? Don't you at least need a metric? For example, can you do calculus in a vector space in which every set is an open set?

For topological spaces X and Y, a function f : X->Y, and points x, x0 in X and y in Y: lim f(x) = y as x->x0 if for every neighborhood V of y, there is a neighborhood U of x0 such that f(U - {x0}) is contained in V. You don't need a metric for that. That's what I meant.