# Tag Info

## Hot answers tagged simulation

26

The only reason I am attempting to answer this is to perhaps get a Reversal badge. There you go... We will go slowly and this answer is the basis for what comes next. Let's start with two dimensions. You'll see why. We create a rectangular region: Needs["NDSolveFEM"] mesh = ToElementMesh[FullRegion[2], {{0, 5}, {0, 1}}, "MeshOrder" -> 1, ...

12

The separation-of-variables solution you quoted has two indices appearing in it: n and j (the subscripts of the coefficient $A_{nj}$). Here, n is azimuthal mode order, i.e. it counts the number of nodes along the direction in which the polar-angle $\theta$ varies (divided by 2). The index j is needed because the wave is supposed to satisfy the boundary ...

8

A couple things: You don't update velocity (obx and oby appear to be unused?). You do your timestep wrong (you want x+=velocity*timestep, not x=velocity*time!). You also have your direction of your force wrong. You should take the negative gradient of potential. The gradient of 1/r is -1/r^2. The negative of that is 1/r^2. So you get the common sense result ...

5

With f[a_, {x_, y_}] := Piecewise[{{a x, x < y}, {x, x == y}, {1 - a + a x, x > y}}, 0] one simulation can be defined by sim[length_] := Module[{rv = RandomVariate[BetaDistribution[3, 1], length], y, yBar}, y[1] = First@rv; yBar[t_Integer] := yBar[t] = 1/t * Sum[y[i], {i, 1, t}]; y[t_Integer] := y[t] = f[0.5, {rv[[t]], yBar[t - 1]}]; ...

1

With the function f[a_, {x_, y_}] := Piecewise[{{a x, x < y}, {x, x == y}, {1 - a + a x, x > y}}, 0] you can define a new function, that will perform a single simulation sim[a_Real, n_Integer] := Module[{data = Partition[Riffle[#, Accumulate[#]/Range[n]], 2] &@ RandomVariate[BetaDistribution[3, 1], n]}, f[a, #] & /@ data ] where the ...

Only top voted, non community-wiki answers of a minimum length are eligible