Print

Please login or register.
1 Hour 1 Day 1 Week 1 Month Forever
Login with username, password and session length

[Graknorke]

Is that the one where you use logarithms to get a linear graph?

[Skyrunner]

What does ∇ · ∇<vector u> mean? Is it the same as divergence(<vector u>)?


edit: one more, what does the big D in Du/Dt (u and t and vectors) mean?

[da_nang]

If ∇·∇u = ∇²u then it's the vector Laplacian. This is a vector and thus can't be the same as the divergence.

The D in Du/Dt is presumably a differential operator. The result in this case would be the Jabocian matrix.

[Skyrunner]

But what's a vector laplacian? @_@ How do you calculate del squared? Is it just the square of the partial derivative?

[MagmaMcFry]

It's a scalar, the sum of all second derivatives of u to each coordinate. It's basically the divergence of the gradient of u.

[MagmaMcFry]

Well, I'm stumped on a little problem.

Given the equation 2ⁿ - k = 3^m - j, where n,m,k,and j are all positive integers (including zero), how would you reduce the function to only being in terms of m and j? In other words, if I have some number that is in terms of powers of 3 and wanted to re-write it in terms of powers of 2, how would I go about this in a reasonable fashion, preferably in an equation format? It is guaranteed that j is not divisible by 3- it doesn't actually matter if k is not divisible by 2, though if that can also be guaranteed, that would be nice.

So given nonnegative integers m and j, you want to find nonnegative integers n and k satisfying the equation? Just set n to 0 or 1 and k to 1+j-3^m or 2+j-3^m depending on parity of 3^m-j. This doesn't seem to be all that useful though. Do you have any more constraints on n and k?

[MagmaMcFry]

Okay, so you basically want the smallest power of 2 containing the number represented by m and j. May I assume that 3^m is the smallest power of 3 containing said number? In this case, n is equal to m multiplied by log(3)/log(2), plus a small number between -2 and 1. So you need to choose the right n from three given values, which is not hard to do algorithmically, and algorithmically is how you need to do this conversion, I assume.

[i2amroy]

*I'm assuming you meant "largest power of 2 that does not contain the number", since that's what you demonstrated in your example.

If that's the case, why not just find the smallest power of 2 that contains the number... and then subtract one from the power?

[da_nang]

Well, I know that n>=m and k<((2^n)-(2^(n-1))).

With the variables being non-negative integers, this implies 2^n-1 + 1 <=  2ⁿ - k <= 2ⁿ and n > 0.
Consequently, 2^n-1 + 1 <=  3^m - j <= 2ⁿ must hold for the equation to have a solution fitting said conditions.
Automatically, one can see that n >= m requires an additional constraint on j since 3^m - j <= 3^m and the condition requires 2ⁿ >= 3^m or equivalently n >= ceil(m log₂(3)).

If we were to choose n = ceil(m log₂(3)), then n >= m is satisfied. No other n can be chosen since there can only exist one m such that 2^n-1 <=  3^m <= 2ⁿ and any other chosen n would result in k not satisfying its condition.
The former is true since otherwise there would exist an integer s > 0 such that 3^{m + s} <= 2ⁿ or equivalently 3^m <= 2ⁿ / 3^s < 2^n-1, which cannot happen since 3^m >= 2^n-1. This is also similarly true for negative s which is omitted here.
Consequently, j must have a value so that 2^{ceil(m log₂(3))-1} + 1 <=  3^m - j <= 2^{ceil(m log₂(3))} or equivalently j < 3^m - 2^{ceil(m log₂(3)) - 1}.

Hence the unique integer solution n = ceil(m log₂(3)) and k = 2ⁿ - 3^m + j when above conditions are met.

Do correct me if I'm wrong.

[Helgoland]

Can anyone imagine what a non-finite CW complex dominated by a finite one would look like? I've been wrecking my brain for days, and I have no idea...

[MagmaMcFry]

Can anyone imagine what a non-finite CW complex dominated by a finite one would look like? I've been wrecking my brain for days, and I have no idea...

What's the definition of "dominated" again?

[Helgoland]

A space X is dominated by a finite CW complex Y if there's maps f and g, X--f-->Y--g-->X with gof homotopic to the identity.

[MagmaMcFry]

Okay, then how about this: Let X = (0,1), Y = [0,1], f : x |-> x, g : y |-> y/2.
Then Y is a finite CW complex, and X is an infinite CW complex, and g°f is homotopic to identity by homotopy H: (x, t) |-> x/(1+t).

[Helgoland]

Ugh, I should've mentioned: The infinite CW complex needs to not be homotopic to a finite one, of course. Else any infinite contractible space would do, yes.

[ZetaX]

Can anyone imagine what a non-finite CW complex dominated by a finite one would look like? I've been wrecking my brain for days, and I have no idea...

The best I can offer is http://en.wikipedia.org/wiki/Wall%27s_finiteness_obstruction . An explicit (but probably to complicated) example is given at the very end of http://www.math.harvard.edu/~lurie/281notes/Lecture2-Wall.pdf .