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/sci/ - Science & Math

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File: 131 KB, 405x349, sierpinski.png [View same] [iqdb] [saucenao] [google]
3879021 No.3879021 [Reply] [Original]

Do you guys want a beautiful problem?

Consider a particle located at the origin. In t "units" the particle will split into 2 particles in which one particle travels to the right, and the other, left, exactly 1 "unit" distance apart. How many particles will exist at t = 129? What about n?

>> No.3879037

what? <span class="math">2^{t}[/spoiler]

>> No.3879058

What kinds of particles? Do overlapping particles count as one? Or do successive overlaps cancel out value?

What about the Pauli Exclusion principle? I mean how many fucking energy states does this particle have?

Get wrecked OP :3

Nice problem though, I have to say.

>> No.3879062

If we count overlapping particles as 1, then the answer is 2^t + 1, by the way.

>> No.3879068
File: 12 KB, 300x100, securetrips.gif [View same] [iqdb] [saucenao] [google]
3879068

>by the way
fag

>> No.3879070

There's many directions to go in this problem, depending on the conditions.

You should make it so that overlaps switch on twice, and then off once.

>> No.3879075

You're right. I don't even know why that's a force of habit for me.

>> No.3879081

that's ok.
i agree that op's question could be better defined.

>> No.3879873

Oh sorry. When they overlap, they cancel each other out.

>> No.3879889

>>3879873

Really looks like a cellular automaton then, with this additional condition. The answer will not be trivial to find and it'll be hard to prove.

>> No.3879893

>>3879021
>can't specify a problem

never change sci

>> No.3879897

>Applied problem
>"Beautiful"

This is what retarded Physics majors actually believe.

>> No.3879898

>>3879889
It's not as hard to prove as you might think. Just start drawing it out, and you'll end up seeing a simple fractal as you expand onto infinity. Furthermore you have probably seen this expansion before in a simple algebra or probability course.

>> No.3879902

>>3879897
I'm not a physics major, I was doing a putnam problem and stumbled onto this.

>> No.3879924

>>3879898
use the fact that c(n,k)=c(n-1,k-1)+c(n-1,k) on pascal's triangle, then reduce all of the coefficients into binary.

1
11
101
1111
10001
etc.

then count each row for the specified time. this is otherwise known as sierpinski gasket or sierpinski's triangle

it's a pretty looking fractal and isn't simply physics

so actually do the problem before u say its retarded

>> No.3879927

>>3879924
but if a problem is retarded, why would you bother doing it?

>> No.3879931

>>3879927
He didn't say it was retarded.

>> No.3879944

>>3879931
> implying I said he did

>> No.3879950

>>3879889
After implementing it (which might help in trying to find a closed form):

If for t=0, there is 1 particle, for t=2 there are 2, for t=3 there are 2, for t=4 there are 4 etc, then for t=129 there are 128.

Here's the sequence of the number of particles at each instant:
1 2 2 4 2 4 4 8 2 4 4 8 4 8 8 16 2 4 4 8 4 8 8 16 4 8 8 16 8 16 16 32 2 4 4 8 4 8 8 16 4 8 8 16 8 16 16 32 4 8 8 16 8 16 16 32 8 16 16 32 16 32 32 64 2 4 4 8 4 8 8 16 4 8 8 16 8 16 16 32 4 8 8 16 8 16 16 32 8 16 16 32 16 32 32 64 4 8 8 16 8 16 16 32 8 16 16 32 16 32 32 64 8 16 16 32 16 32 32 64 16 32 32 64 32 64 64 128

Finding a pattern might be doable indeed. Must be related to the binary representation of t-1 I think.

>> No.3879971

>>3879950
hmmm... I think I remember coming across this pattern in a project euler problem.

>> No.3879975

>>3879971
http://oeis.org/A001316

>> No.3879983

>>3879950
Ok, so indeed, the number of particles at time <span class="math">t[/spoiler] can be written as <span class="math">2^{w(t-1)}[/spoiler] where <span class="math">w(t-1)[/spoiler] is the number of <span class="math">1[/spoiler]'s in the binary expression of <span class="math">t-1[/spoiler].

I don't have a proof right now though. I guess induction probably works?

>> No.3879987
File: 79 KB, 936x751, splitting_particles.png [View same] [iqdb] [saucenao] [google]
3879987

>>3879983
Forgot to add picture of code (left) and results (right). Results are decimal expression of t-1, next to numbers of particle at time t.

>> No.3879989

Havent thought about the problem, but from the sequence it seems evident that every <span class="math">2^n[/spoiler] steps you have the full <span class="math">2^n[/spoiler] particles. Should be quite possible to prove that by induciton.

>> No.3879990

>>3879971
Yeah, it's very "projecteuler-like". I think I've done that exercise too.

>> No.3879997

I presume the length of a space unit at each time unit is equal to the space unit used in the last time unit divided by the square root of two. There will be n to the power of 2 particles at every positive stage. They will never overlap and their path will form an H-Tree.

>> No.3880087

>>3879997
You're right. Now I feel stupid I didn't presume the same, it would have made perfect sense.

>> No.3880097

>>3879997
> n to the power of 2
lol

>> No.3880099

>>3880087
sorry for off topic but can you post the link to your version of the previewer again?

>> No.3880164

>>3880099
There it is:
http://userscripts.org/scripts/show/114958

However, it doesn't work with the QR box on the main pages. It works with 4chan's boxes in both specific threads and in the front page, but if you use the quick reply box from 4chanX's script, then you need to do it from within a thread.