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Topic: Three Dimensional Random Walk (Read 11649 times) |
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Leo Broukhis
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Re: Three Dimensional Random Walk
« Reply #25 on: Jun 11th, 2003, 5:48pm » |
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I ran a simulation of 10000 points for 100000 steps four times and got 28.27%, 28.11%, 27.42%, and 28.57%.
Another run of 100000 points for 100000 steps ended with 28163 points returning.
JFYI: The same run in 2 dimensions results in 78024 points returning.
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| « Last Edit: Jun 11th, 2003, 6:30pm by Leo Broukhis » |
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James Fingas
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Found Recurrance Relation! w0000007!
« Reply #26 on: Jun 12th, 2003, 6:05am » |
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Haha! I have found a way to make this problem mathematically tractable!
2) After two moves, exactly 1/8 of the particles have returned to the origin. This accounts for 8 out of 64 particles. This gives a fraction of 0.125.
4) After four moves, there are (4C2)3=63=216 particles at the origin. Some of these are returning from visiting the origin on move 2. But we know exactly how many are doing this: 8 for each particle that was at the origin on move 2, for 64 particles total. Therefore 216-64=152 particles are returning for the first time. This gives a fraction of ~0.03711.
6) After six moves, there are (6C3)3=203=8000 particles at the origin. Some of these first visited the origin on move 2, and some first visited the origin on move 4. But we know exactly how many: 8*216 first visited on the second move, and 152*8 first visited on the fourth move. This leaves 8000-8*216-152*8=8000-1728-1216=5056 particles returning for the first time. This gives a fraction of ~0.019287.
8) After eight moves, there are (8C4)3=703=343000 particles at the origin. Some of these first visited on moves 2,4, or 6. But again we know exactly how many: 8*8000=64000, 152*216=32832, and 5056*8=40448 respectively. This means 343000-64000-32832-40448=205720 particles total return to the origin for the first time. This gives a fraction of ~0.0122619.
This can be extended to an arbitrary number of steps, and takes O(N2) computational power. Working this out on a spreadsheet my numbers exactly match towr's! This method allows us to come up with a recurrance relation for the number of particles that eventually return to the origin.
Let us define M(N) = (2NCN)3. Now define the sequence Q(N) which gives the number of particles that return to the origin for the first time on step 2N:
Q(N) = M(N) - M(N-1)Q(1) - M(N-2)Q(2) - M(N-3)Q(3) - M(N-4)Q(4) - ... - M(1)Q(N-1)
It may be possible to solve the recurrance relation explicity for Q(N), or it may be possible to calculate the probability without doing that:
P = sumn=1..inf Q(n)/64n
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ThudnBlunder
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Re: Three Dimensional Random Walk
« Reply #27 on: Jun 13th, 2003, 7:39pm » |
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To further confuse things, if my (and Leonid's) answer includes not only points that arrive back simultaneously once from 3 independent 1-D walks, but also counts ones that arrive back together again 2, 3, 4, 5...... times, then don't we just have an infinite geometric series whose sum is mine and Leonid's answer?
That is, x/(1-x) = 0.3126526...
This gives x = 0.2381838... 
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| « Last Edit: Jun 13th, 2003, 11:03pm by ThudnBlunder » |
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James Fingas
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Re: Three Dimensional Random Walk
« Reply #28 on: Jun 16th, 2003, 1:45pm » |
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T&B,
Looking at towr's code, I'm pretty sure it gives correct numbers. I don't know about your simulation though. The problem with this one is that you need to remove those particle paths that return to the origin; otherwise, they'll return again and again and spoil your results.
This is easy to do in a simulation, but hard to do in the math. My method with Q and M works (or reproduces towr's results anyways). A straight geometric series can't do the trick because the particles can return once or twice or a thousand times, and the probabilities are all based on the lengths of time between the returns to zero. You can calculate the probabilities directly, but it's O(N3) and rather complex, versus O(N2) for my method. To see what I mean, expand my method in terms of just the Ms: it gets hairy pretty fast!
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Leo Broukhis
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Re: Three Dimensional Random Walk
« Reply #29 on: Jun 16th, 2003, 2:51pm » |
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on Jun 13th, 2003, 7:39pm, THUDandBLUNDER wrote:To further confuse things, if my (and Leonid's) answer includes not only points that arrive back simultaneously once from 3 independent 1-D walks, but also counts ones that arrive back together again 2, 3, 4, 5...... times, then don't we just have an infinite geometric series whose sum is mine and Leonid's answer?
That is, x/(1-x) = 0.3126526...
This gives x = 0.2381838...
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No, my simulation drops a point from the simulated set as soon as it reaches the origin.
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Barukh
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Re: Three Dimensional Random Walk
« Reply #30 on: Aug 20th, 2003, 5:09am » |
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There are quite a few sites talking about this problem, for example: http://www.mathsoft.com/mathresources/constants/discretestructures/artic le/0,,2220,00.html.
It says that the sought probability is 0.340537...
Let P(D) be the probability for the particle to get back in D-dimensional space, and Q(D) = 1 - P(D). Let also pN be the probability that the 1-dimensional walk gets to the origin after 2N steps (not necessarily for the first time!). Then, the particle does not get to the origin in the D-dimensional space if it doesn't get to the origin after any finite number of steps. The probability not to be at the origin after 2N steps is 1 - pND, so
Q(D) = 1PRODoo (1 - pND).
We know that pN = 2NCN/22N. Thus, pN+1/pN = (2N-1)/2N. So, computationally, it takes O(N) time and constant space. The simulations show the following numbers:
N P(3)
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10^3 0.32487
10^4 0.33010
10^5 0.33174
10^6 0.33226
10^7 0.33243
10^8 0.33248
10^9 0.33249
Although the numbers are quite close to the theoretical value, the approaching rate is very slow... Or, there is an error in reasoning?..
That the Leonid's and THUDandBLUNDER's formula is not correct, is seen from the following considerations: the formula should work for any number of dimensions, only powers change. However, for D=1 and D=2 the infiinite sum diverges and cannot be a probability. But it is handy in the following approach (taken from a nice little Mosteller's book `Fifty Challenging Problems in Probability').
Let XN be a random variable that takes value 1 if the particle is at the origin after 2N steps, and 0 otherwise, and let X = 1SUMoo XN. The expected value E[XN] = pND, so E[X] is just the Leonid's and THUDandBLUNDER's formula!
It is easy to see that E[X] is an expected value of returns until particle is lost forever. Now, we may associate the random walk with flipping of the biased coin, where P(D) is the probability to get heads, Q(D) is the probability to get tails. Then, 1 + E[X] is an expected number of flips until the first tails. So, Q(D) = 1/(1 + E[X]). The simulation shows the following numbers:
N P(3)
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10^3 0.27633
10^4 0.28037
10^5 0.28164
10^6 0.28204
10^7 0.28217
10^8 0.28221
10^9 0.28222
Hmm, the numbers are much lower than the theoretical value. Is this because of the program bug? convergence rate? floating point errors? Whatever it is, I cannot see the flaw in the reasoning...
This approach also clearly shows why P(D) is 1 for D=1, 2 and <1 for higher dimensions. It is clear that P(D) = 1 iff E[X] = oo. Using Stirling's formula, one gets that E[XN] ~ N-D/2, and the sum diverges only for D=1, 2.
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James Fingas
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Re: Three Dimensional Random Walk
« Reply #31 on: Aug 20th, 2003, 6:57am » |
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Barukh,
I think the reason your formula
Q(D) = [prod]N=1..[infty](1-pND)
does not converge to the correct answer is that it assumes that the probability of a particle returning on step N is independent of the probability that it returned on step N-1. If a particle doesn't return by step N it is less likely than average to return by step N+1. However, I would have said that because of this, you would predict an answer lower than the actual answer...
Please note that the answer 0.340537... is probably not relevant to this question, because it assumes you move in one of six directions (+x, -x, +y, -y, +z, -z), while we are allowed to move in eight directions (+++, ++-, +-+, +--, -++, -+-, --+, ---). This was noted by TnB a few posts up the page. He supposed that our answer would be less than 0.340537 because of the larger 'branching factor'.
Our best guess at the probability is from simulation, giving at least 28%, but not likely as high as 34%.
Your criticism of Leonid's and TnB's answer could also be applied to your own. Although you predict that Q(1) = 0 and Q(2) = 0, I am quite confident that the number of particles that have never returned to the origin by step 2N for a 1D random walk is 2NCN, which is significantly larger than your answer.
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Barukh
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Re: Three Dimensional Random Walk
« Reply #32 on: Aug 20th, 2003, 10:44am » |
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James,
on Aug 20th, 2003, 6:57am, James Fingas wrote:Barukh,
I think the reason your formula
Q(D) = [prod]N=1..[infty](1-pND)
does not converge to the correct answer is that it assumes that the probability of a particle returning on step N is independent of the probability that it returned on step N-1. If a particle doesn't return by step N it is less likely than average to return by step N+1. However, I would have said that because of this, you would predict an answer lower than the actual answer... |
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You are right... I didn't think about this dependency...
on Aug 20th, 2003, 6:57am, James Fingas wrote:
Please note that the answer 0.340537... is probably not relevant to this question, because it assumes you move in one of six directions (+x, -x, +y, -y, +z, -z), while we are allowed to move in eight directions (+++, ++-, +-+, +--, -++, -+-, --+, ---). This was noted by TnB a few posts up the page. He supposed that our answer would be less than 0.340537 because of the larger 'branching factor'.
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Agreed. I overlooked this.
on Aug 20th, 2003, 6:57am, James Fingas wrote:
Our best guess at the probability is from simulation, giving at least 28%, but not likely as high as 34%. |
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Still, I think the second approach is correct.
on Aug 20th, 2003, 6:57am, James Fingas wrote:
Your criticism of Leonid's and TnB's answer could also be applied to your own. Although you predict that Q(1) = 0 and Q(2) = 0, I am quite confident that the number of particles that have never returned to the origin by step 2N for a 1D random walk is 2NCN, which is significantly larger than your answer. |
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Sorry, I don't understand this.
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towr
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Re: Three Dimensional Random Walk
« Reply #33 on: Aug 20th, 2003, 10:55am » |
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on Aug 20th, 2003, 10:44am, Barukh wrote:
Still, I think the second approach is correct. |
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That's the one which goes with the second table, right? which gives the +- 28% we expect. So I also think that's correct.
How did you simulate that? I tried to calculate all the probabilities and add them, but the rather large factorials just take too much time..
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James Fingas
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Re: Three Dimensional Random Walk
« Reply #34 on: Aug 20th, 2003, 11:19am » |
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towr,
You're talking about the problem with calculating pN right? The simplest way to do this is with the recurrance relation pN=pN-1*(2N-1)/2N.
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towr
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Re: Three Dimensional Random Walk
« Reply #35 on: Aug 20th, 2003, 11:29am » |
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hmm.. yes.. I should have read his post more carefully, it says so right there..
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James Fingas
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Re: Three Dimensional Random Walk
« Reply #36 on: Aug 20th, 2003, 12:59pm » |
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on Aug 20th, 2003, 5:09am, Barukh wrote:| It is easy to see that E[X] is an expected value of returns until particle is lost forever. Now, we may associate the random walk with flipping of the biased coin, where P(D) is the probability to get heads, Q(D) is the probability to get tails. Then, 1 + E[X] is an expected number of flips until the first tails. So, Q(D) = 1/(1 + E[X]). The simulation shows the following numbers: |
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Essentially, we are saying there is a well-defined probability Q(D) that a particle currently at the origin will never return. Whenever a particle departs from the origin, we flip a coin calculating whether or not it will ever come back. In this way, we get a geometrically-decreasing percentage of particles that keep coming back.
Particles return with probability P(D).
0: 1
1: P(D)
2: P(D)2
3: P(D)3
...
But E[X] manages to count all of these (except for N=0), so we just need to add 1 to E[X] and work backwards to find P(D).
Hmm ... it seems valid ... good work!
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Sjoerd Job Postmus
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Re: Three Dimensional Random Walk
« Reply #37 on: Mar 21st, 2005, 6:54am » |
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I'll try and trow myself at this question.
Let me first analyse the 1D situation.
When I make 2x moves, exactly x of those need to go left, and the other x to go right.
if (x%2) -> FALSE, can not be done
if (!(x%2) -> Maybe...
When I make 2n moves, I have (2n)! / (n!)^2 ways to reach the center. (From 2n moves, I choose n rights) I have 2^(2n) different options. So the probability for n:
(2n)! / ((n!)^2 * 2^(2n))
Now, if I do this in 3D, I would just have to multiply the probabilities.
For n moves, the prob. would be:
((2n)! / ((n!)^2 * 2^(2n)))^3
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Deedlit
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Re: Three Dimensional Random Walk
« Reply #38 on: Mar 27th, 2005, 3:39am » |
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I haven't read through all the replies, but for what it's worth here is one answer:
The desired probability is the probability that three simple random walks return to the x-axis simultaneously. To return to the x-axis for the first time at y=2n+2, the walk goes in one direction at the first step, stays on the same side and returns to +/-1 at 2n+1, and goes to zero at 2n+2. The middle part is equivalent to a nonnegative walk that returns to zero at 2n, and there are Catalan(n) = C(2n,n)/(n+1) ways of doing so. So the probability of returning at 2n+2 is
P(2n+2) = 2 * [C(2n,n)/(n+1)] / 22n+2
So the required answer is
Sumn=0inf P(2n+2)3 = Sumn=0inf C(2n,n)3/[(n+1)3 26n+3]
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Barukh
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Re: Three Dimensional Random Walk
« Reply #39 on: Mar 27th, 2005, 5:55am » |
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Deedlit, I think there is a problem with your formula: P(n) gives the probability of 3 independent processes returning to the origin after n steps simultaneously for the first time.
While the correct answer will take into account the fact that the first return requires at least one dimension to return for the first time, while other directions may return an arbitrary number of times. This is the point that makes the computation of the probability so difficult.
You may consider looking at my post where the method using indicator random variables is presented. As far I know, there was a consensus that this method gives the correct answer.
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| « Last Edit: Mar 27th, 2005, 6:02am by Barukh » |
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Deedlit
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Re: Three Dimensional Random Walk
« Reply #40 on: Mar 28th, 2005, 7:25pm » |
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Whoops! You're right of course. Using the expected value to find the probability looks right - I'm amazed I didn't know this already. I probably learned it before and forgot about it. 
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rmsgrey
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Re: Three Dimensional Random Walk
« Reply #41 on: Apr 4th, 2005, 6:22am » |
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on Mar 27th, 2005, 5:55am, Barukh wrote:| While the correct answer will take into account the fact that the first return requires at least one dimension to return for the first time, while other directions may return an arbitrary number of times. This is the point that makes the computation of the probability so difficult. |
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Ummm... Why can't all 3 co-ordinates return to 0 multiple times before the first return to (0,0,0)?
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SWF
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Re: Three Dimensional Random Walk
« Reply #42 on: Apr 5th, 2005, 5:48pm » |
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The condition that a jump be made in each of three direction for every time increment, makes this equivalent to random jumps to nearest neighbors on a body centered cubic lattice.
The probability is found to be 1 - [pi]2/(4*A2) where
A = the integral from 0 to [pi]/2 of dt/sqrt(1 - (1/2)*sin2t)
(The complete elliptic integral of the first kind for 1/sqrt(2)). The probability works out to be 0.282229989...
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| « Last Edit: Apr 5th, 2005, 5:49pm by SWF » |
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Deedlit
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Re: Three Dimensional Random Walk
« Reply #43 on: Apr 6th, 2005, 4:58pm » |
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Nice - could you fill in some more details?
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SWF
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Re: Three Dimensional Random Walk
« Reply #44 on: Apr 12th, 2005, 7:26pm » |
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Here is a summary of the method I used, which should work with all sorts of random walks in various types of lattices of any dimension. The same approach can also be used to find resistance between points in an array of resistors. Start with an N by N by N array of points (assume N is an even number). Let p(m,n,q) be the probability that a random walk starting at point (m,n,q) eventually arrives at point (N/2,N/2,N/2).
A point already there is certain to reach that point, so p(N/2,N/2,N/2) = 1. For motion with the given rules:
8*p(m,n,q) - p(m[pm]1,n[pm]1,q[pm]1])=f(m,n,q). The term with [pm] means plus or minus and represents 8 different terms, and f(m,n,q) equals zero for every (m,n,q) except for (N/2,N/2,N/2) where it equals some constant, B.
Use discrete Fourier Transform: G(w)=sum(g(j)*exp(-2[pi]i*w*j/N), where sum is on j from 0 to N-1. Take discrete Fourier Transform of the expression for f(m,n,q) three times summing over m, n, and q. Some algebra leaves the transform of p(m,n,q):
P(a,b,c)= B*exp(-[pi]i(a+b+c))/8/(1-cos(2[pi]a/N)*cos(2[pi]b/N)*cos(2[pi]c/N)
The formula for an inverse discrete Fourier Transform is g(j)=(1/N)*sum(G(w)*exp(2[pi]i*w*j/N), where the sum is on w from 0 to N-1. To find p(m,n,q), take the inverse transform on the three parameters, a, b, and c. Take limit on the triple sum as N goes to infinity to get a triple integral.
Using the resulting formula gives p(N/2,N/2,N/2)= (B/64/[pi]3)*[int] dx*dy*dz/(1-cos(x)*cos(y)*cos(z)), where "[int]" means triple integral on x, y, and z from 0 to 2[pi]. Knowing that p(N/2,N/2,N/2) equals 1 and evaluating the integral, allows the constant B to be evaluated. Also by its definition, B = 8 - 8*p(N/2+1,N/2+1,N/2+1). This allows evaluation of p(N/2+1,N/2+1,N/2+1), the probability of a point one step away from the starting point to ever return.
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Deedlit
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Re: Three Dimensional Random Walk
« Reply #45 on: Apr 13th, 2005, 9:01pm » |
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on Apr 12th, 2005, 7:26pm, SWF wrote:
Use discrete Fourier Transform: G(w)=sum(g(j)*exp(-2[pi]i*w*j/N), where sum is on j from 0 to N-1. Take discrete Fourier Transform of the expression for f(m,n,q) three times summing over m, n, and q. Some algebra leaves the transform of p(m,n,q):
P(a,b,c)= B*exp(-[pi]i(a+b+c))/8/(1-cos(2[pi]a/N)*cos(2[pi]b/N)*cos(2[pi]c/N)
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How did you get that? If (N/2, N/2, N/2) is the only nonzero value, then I only get one term from the fourier transorm, B*exp(-[pi]i(a+b+c)). Do the other terms come from boundary cases?
Quote:| The formula for an inverse discrete Fourier Transform is g(j)=(1/N)*sum(G(w)*exp(2[pi]i*w*j/N), where the sum is on w from 0 to N-1. To find p(m,n,q), take the inverse transform on the three parameters, a, b, and c. Take limit on the triple sum as N goes to infinity to get a triple integral. |
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I'm not there yet, but what is the rationale for taking the transform and then immediately the inverse transform?
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SWF
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Re: Three Dimensional Random Walk
« Reply #46 on: Apr 14th, 2005, 9:35pm » |
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You do not take the transform and immediately invert it. Take the transform of the difference equation. The transform of f(m,n,q) is equal to B*exp(-[pi]i(a+b+c)) and that equals the transform of
8*p(m,n,q) - p(m-1,n-1,q-1) - p(m-1,n-1,q+1) - ... - p(m+1,n+1,q+1)
which is P(a,b,c)*8*(1-cos(2[pi]a/N)*cos(2[pi]b/N)*cos(2[pi]c/N)). Rearrage to get an expression for P(a,b,c) (the transform of p(m,n,q)) then take the inverse. The whole trick is that the transform of that difference equation leads to something which can be inverted to get p(m,n,q).
When transforming that difference equation to get the above result, you have to assume that p() is periodic with period N and manipulate the limits on the summations so that each of the 9 terms is clearly a multiple of P(a,b,c). That leads to a bunch of exp() of imaginary numbers, and use 2*cos(x)=exp(i*x)+exp(-i*x).
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Deedlit
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Re: Three Dimensional Random Walk
« Reply #47 on: Apr 15th, 2005, 3:40pm » |
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Ah, I see. Just a couple more questions, if you don't mind: How did you evaluate the integral, and in what sort of situations do you use the Fourier transform?
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SWF
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Re: Three Dimensional Random Walk
« Reply #48 on: Apr 20th, 2005, 5:38pm » |
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I don't remember the details, but here is a rough summary of how a value was found for that integral. The first of the triple integrals was done with residues, so that it becomes a double integral with an integrand of something like 2[pi]/sqrt(1-cos2(y)*cos2(z)). This is proportional to the integral of the complete elliptic integral of first kind of cos(y). Change of variable makes it proportional to integral from 0 to 1 of K(w)/sqrt(1-w2), where K(w) is complete elliptic integral of first kind of w. That last integral looked tough, but checking in a book of integral tables, it is found to equal K(1/sqrt(2))2.
Fourier Transform is commonly used for spectral analysis of a signal. It transforms data into their frequency components. Electrical engineers, such as William Wu, can explain better than I can.
Fourier Transform is also useful for solving certain partial differential equations, such as Laplace Equation in rectangular coordinates, when the domain is -infinity to +infinity. Similarity to the difference equation in this problem to the Laplace equation is why I considered Fourier Transform. For solving a PDE an integral form of the FT is used, so I figured a discrete version could be used for a difference equation.
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River Phoenix
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Re: Three Dimensional Random Walk
« Reply #49 on: May 22nd, 2005, 1:47pm » |
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Can this be simplified to a 1-dimensional random walk, since a particle which is, say, 7 steps from the origin, is symmetrical with every other unit which is 7 steps from the origin. Then it can probably be solved by a recurrence relationship?
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