I don't think so, because there is no sensible generalization of "vector space" or "orthonormal basis" to dimensions other than whole numbers; more generally, IMO a set of non-integral dimension cannot even be a manifold.
I wish psxjin had the TAS ISO plugin that allowed you to make .bz archives from the ISOs and then load them, in order to save space.
This will never happen, ever.
Expect miserable failure in about a week.
Did adelikat merely mean that PSXjin will never have that plugin (the release announcement said "no plugins") or that it will never have the features that the plugin gave to PCSX-rr?
IDK whether they have an API, but they do have a widget called Google Translate Web Element that can be placed on a webpage so that users can decide whether they want it translated: https://translate.google.com/support/?hl=en
For some reason Mupen64 crashes during video capture when it reaches a part in the ending credits.
It's possible for me to do an encode leaving the credits out of it but it's prolly easier to let someone else that don't get this crash give it a go.
IMO you should produce an encode without the credits, at least for distribution via YouTube.
I expected someone to get tripped up by saying "stage 3, because that's when you'd have a 50/50 chance of getting an Item"; similarly I'll try to figure out the expected stages by which 2 Items and all 3 Items would have been obtained...
2 Items
2: 3/8*2/7=3/28, about 11%
3: 3/28+2*5/8*3/7*2/6=2/7, about 29%
4: 2/7+3*5/8*4/7*3/6*2/5=1/2, 50%
all 3 Items
3: 3/8*2/7*1/6=1/56, about 2%
4: 1/56+3*5/8*3/7*2/6*1/5=1/14, about 7%
5: 1/14+6*5/8*4/7*3/6*2/5*1/4=5/28, about 18%
6: 5/28+10*5/8*4/7*3/6*3/5*2/4*1/3=5/14, about 36%
7: 5/14+15*5/8*4/7*3/6*2/5*3/4*2/3*1/2=35/56, 62.5%
more generally...
If there are M>0 items scattered among N>M stages, the probability of having found at least K in [0,M] of them by stage J in [K,N] is
SUM(C(I-1,K-1)*P(N-M,I-K)*K!/P(N,N-I),I,K,J)
For those unfamiliar with the game, Mega Man 2 starts with 8 stages that can be completed in any order, and 3 of them yield Items; if you choose the stages randomly, by the completion of which stage are you more likely than not to have at least one of the three Items?
It's simple to compute Fibonacci and Lucas numbers using only integer math, by taking advantage of the following facts:
L(A+B)=(L(A)L(B)+5F(A)F(B))/2
F(A+B)=(L(A)F(B)+L(B)F(A))/2
L(2A)=(L(A)^2+5F(A)^2)/2
F(2A)=L(A)F(A)
I have an old program which computes the Nth Fibonacci number and the Nth Lucas number at once using the above relations, for N up to 46999, which takes about 60ms to compute.
I have been inspired to make a program in Javascript that does this for any number, by first finding its base-2 representation; here's the basic idea:
Set up arrays F[] and L[], remembering that F[1]=1 and L[1]=1, and use the special-case logic that F(0)=0, L(0)=2, F(-N)=(-1)^(N+1)*F(N), and L(-N)=(-1)^N*L(N).
Set up array D[] as the set of powers of 2 to use, something like this...
for(i=0,j=N;j>0;i++){
D[i]=j%2;
j=(j-D[i])/2;
}
Then populate F[] and L[] with values for the powers of 2 up to 2^(D.length()-1) and use the sum formulas to calculate F[N] and L[N].
I'll link to it when I'm done, but in the meantime check out a demonstration of the Huntington-Hill algorithm for U.S. Congressional apportionment: http://jansal.net/HuntingtonHill.shtml
Perfection is an ideal rather than a reality we claim to attain, for we lack perfect insight into the emergent properties of the games and the limitations of our emulators.
While we're on the subject of differential equations, here's a problem I came across...
Consider the following differential equation in 3 dimensions. D2 is the Laplacian operator, and x is the coordinate vector.
- D2f(x) + |x|2 f(x) = (2n + 3) f(x)
Those who know quantum mechanics will recognize this as a modified version of Schrodinger's equation for the isotropic quantum harmonic oscillator. And, in particular, that it only has square integrable solutions for nonnegative n.
However, f(x) = |x|-1 e-|x|2/2 is square integrable.
-Inf/Inf f(x)2 d3x = 0/Inf-1/10/2 pi |x|-2 e-|x|2 |x|2 dp d cos(t) d|x| = 4 pi 0/Inf e-x2 dx = 2 pi3/2
Now try plugging it into the equation and see what you get. Seems like a solution with n = -1, doesn't it? It's not. Why not?
It was a collaboration project after all. Made with a bunch of small tiles. (That image is just a small 5x5 section.) I remember a separate site which would arrange them randomly and you could pan it around.
It was all started by SomethingAwful, it was a "Blue Ball Machine"
I saw this problem on a math forum, and for some reason it got deleted...
Draw two circles that intersect in two distinct points, and draw one of the common tangents and then the circle through both points of tangency and either point of intersection of the circles; prove that the radius of this third circle depends only on the radii of the first two, and not on their position.
I tried using coordinate geometry but ended up with some nested radicals or a quartic or something else scary.
Hey, how do I watch the tool assisted run Calculus posted? I don't seem to have anything that can run it.
If you meant "criticaluser" then you need a copy of the Advance Wars (U) ROM (don't ask us where to find it) and also VBA-rr v23: http://code.google.com/p/vba-rerecording/downloads/list
Open the ROM (unzip or unrar it if necessary, you should end up with a .gba file) in VBA-rr and then I think there's something in the File menu that allows you to play the input file (the .vbm)
Expect miserable failure in about a week.