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2

This is a "computer assisted" solution rather than certified closed form but it may satisfy your needs. F1[n_, t_] := F1[n, t] = If[n >= 1, F1[n - 1, 2 t], t]; Table[F1[n, t], {n, 0, 10}] FindSequenceFunction[%] output : {t, 2 t, 4 t, 8 t, 16 t, 32 t, 64 t, 128 t, 256 t, 512 t, 1024 t} 2^(-1 + #1) t & and F2[n_, y_, t_] := F2[n, y, t] = If[n ...

1

You could also use: cm[f_, s_, t_, n_] := InverseLaplaceTransform[LaplaceTransform[f, s, t]^(n + 1), t, s] (s>0): e.g. TableForm[Table[{j, cm[Exp[-a s] UnitStep[s], s, t, j]}, {j, 1, 10}], TableHeadings -> {None, {"n", "n-fold Convolution"}}]

0

Thanks to all who helped or tried to help me here, finally I found a solution. My special function was the exponential distribution but one can apply it for any arbitrary function. f[x_] = \[Lambda]*Exp[-\[Lambda]*x] UnitStep[x]; g[x_] = \[Lambda]*Exp[-\[Lambda]*x] UnitStep[x]; convi[n_] := {Do[ g[x[i]] = Convolve[f[x[i - 1]], g[x[i - 1]], x[i - 1], ...

4

Here is a hint. Consider you want to implement a function $f(n)=2^n$ using NetList. You will do something like this: NestList[#*2&,1,5] So we used the fact that $f(n+1)=2f(n)$. Now we will implement Fibonacci sequence where $g(n)=g(n-1)+g(n-2)$ NestList[{#[[2]],#[[1]]+#[[2]]}&,{0,1},10] ...

6

Several changes are required to obtain the desired results. First, the syntax error mu[n] == mu must be replaced by mu[n] = mu. Next, initial conditions must be provided for the recurrence in n: mu[0] := mu /. FindRoot[f[1/2, mu, 2^0] == 1/2, {mu, 0.9}] mu[1] := mu /. FindRoot[f[1/2, mu, 2^1] == 1/2, {mu, 0.9}] mu[2] := mu /. FindRoot[f[1/2, mu, 2^2] == ...

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