Recent content by pig

Are you sure this is what is taking a long time? I use MPI in C, not Fortran, but what is slow in a MPI program is starting all the processes. This can take up to a minute sometimes when there's a lot of them. Try adding a Barrier before the Send loop, and after the barrier print something on...

I don't know anything about the implementation you are using so I don't know whether your code is correct, but from my experience such problems are usually the result of using different padding schemes. Also, since it's AES, make sure you are using the same mode (ECB, CBC, CTR...) and...

I somehow managed to not notice that link even though I've read the wiki page. Thanks a lot!

Hey, Is there a proof of Kirchoff's theorem available somewhere online? In literature I can find only proofs of Cayley's formula, and the matrix tree theorem is usually only mentioned, if that. I want to see a proof because I'm having a hard time intuitively understanding why this works...

It's not that mathematicians are more likely to make mistakes, but people who didn't study anything math-related tend to think it's a big deal that a mathematician made such a mistake so it gets noticed more.

This reminded me of my high school math teacher, she wrote "DERIVATIVES" in big letters (well, not in english) on the board and the derivatives of elementary functions, the (f*g)' = ... and the other couple of rules underneath. Then she showed us how to calculate them. She made no mention of...

5 + 4x - x^2 = m - (x -n)^2 5 + 4x - x^2 = m - x^2 + 2nx - n^2 4x - x^2 + x^2 - 2nx = m - 5 - n^2 4x - 2nx = m - 5 - n^2 x(4 - 2n) = m - 5 - n^2 In order for this to work for all values of x, it must not depend on x. And it won't depend on x if 4 - 2n = 0, because x * 0 will be 0...

Yes, and x can also be 2n.

(x - n)^2 = (x - n)(x - n) = (x*x - x*n - n*x + n*n) = x^2 - 2nx + n^2 So, m - (x-n)^2 = m - (x^2 - 2nx + n^2) = m - x^2 + 2nx - n^2

Philocrat, how does logic not apply to the real world?

Ok I made some progress but it still isn't complete: (a over b means the binominal coefficient) (2n!)/[n!(n+1)!] = (2n)!/[n!n!(n+1)] = (2n)!(n+1-n)/[n!n!(n+1)] = [(2n!)(n+1) - (2n!)n]/[n!n!(n+1)] = (2n!)(n+1)/[n!n!(n+1)] - (2n!)n/[n!n!(n+1)] = (2n!)/(n!n!) - (2n!)/[(n-1)!(n+1)!] =...

There are two simple ways to substitute. First, you can add 1 to both sides of the second equation: 2y + 1 = x - 1 + 1 2y + 1 = x So now you can substitute 2y+1 instead of x in the first equation. Another way would be to multiply both sides with 1/2: 2y * 1/2 = (x - 1) * 1/2 y =...

You can use the Euler formula: cos(x) + isin(x) = e^(ix) So: cos(x+y) + isin(x+y) = e^[i(x+y)] cos(x+y) + isin(x+y) = e^(ix+iy) cos(x+y) + isin(x+y) = e^ix * e^iy Turn the right side into sines and cosines using the original formula and see what you get. This way you can prove the...

This also looks wrong to me, but it's too late now, I'll look more closely tomorrow.

Could you draw a picture? I got a different formula, but I'm not sure I understand the problem... The way I understand it, you have a half-cylinder tank, and the large flat part is facing up? Wouldn't then dV be 8*2sqrt(4-y^2)dy? Where does the y in 4y come from? And since distance is y+1...