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[RedWarrior0]

So when one non-constant width figure in another non-constant width figure is maximal but can rotate.



Anyways, I have an exam tomorrow and am having trouble with Stokes' theorem (Specifically in 3 dimensional space). Anyone who can help?

[Vector]

What kind of help do you need?

[RedWarrior0]

Never mind, I think I've got it now, and I should probably be going to bed now anyways.

[Brilliand]

So when one non-constant width figure in another non-constant width figure is maximal but can rotate.

It isn't necessary for a figure to be fully constant-width for it to be able to rotate when maximal.  It's sufficient for it to be constant-width over that small portion of its rotation that happens to be maximal.

Also, for any shape, it's possible to construct a container shape that will allow it to rotate when maximal.  A donut does this for every shape.  Your rule is probably true when all of the figures involved are composed solely of straight lines, though.

[Skyrunner]

lim[n->0, (1+1/n)^n]=1
Why? O_o Is the way to show that the above equation is true too complex for someone who's only done half of calculus introduction(the part with differentiation)?
I asked my math teacher and he thought about it for a while then gave up after a few false starts. Just curious.

[MagmaMcFry]

(1+1/n) converges against 1, and f(n)^n converges against 1 for positively convergent f(n), so (1+1/n)^n converges against 1.

[Skyrunner]

Doesn't (1+1/n) diverge in this case?

[Mego]

Doesn't (1+1/n) diverge in this case?

Nope. a/infinity (for any a) is 0. 1 + 1/n [n -> infinity] = 1 + 0 = 1.

[Skyrunner]

I sent n to 0 though. Look again, it's n->0 and 1/0 goes to infinity.

[MagmaMcFry]

Doesn't (1+1/n) diverge in this case?

Er, yeah. Hm.

EDIT: I got it. If you replace n by 1/x, you get lim[x->infinity, (x+1)^(1/x)], which obviously converges against 1, since x+1 grows smaller than a^x for all a > 1.

[Another]

I think you can see it if you do lim[n->0, (1+1/n)^n]=lim[n->0, e^ln((1+1/n)^n)]=lim[n->0, e^(ln(1+1/n)*n)] and notice, that ln(1+x)/x goes to zero as x goes to +inf.

[ed boy]

lim[n->0, (1+1/n)^n]=1
Why? O_o Is the way to show that the above equation is true too complex for someone who's only done half of calculus introduction(the part with differentiation)?
I asked my math teacher and he thought about it for a while then gave up after a few false starts. Just curious.

If it exists, lim[n->0, (1+1/n)^n]=lim[n->infinity, (1+n)^(1/n)].
You can show that this is strictly decreasing for n>0 (take the derivative if needs be), and that it is bounded below by 0 for n>0 (as 1+n is positive). Therefore, it converges to a limit.

[palsch]

X^0 = 1 for all X. Therefore lim[n->0 f(n)^n]=1 for all f(n). The nature of the convergence depends on f(n), but the limit itself doesn't.

[MagmaMcFry]

X^0 = 1 for all X. Therefore lim[n->0 f(n)^n]=1 for all f(n). The nature of the convergence depends on f(n), but the limit itself doesn't.

That's exactly what a limit value is not. The limit value of a function at a point does not at all depend on the value of the function at that point. For example, take the function z(x), which is 1 when x is 0, but 0 otherwise. Then lim[x->0, z(x)] is 0, not 1.

[palsch]

Disregard, half asleep. Will try to decode what I meant when I have time.