New answers tagged difference-equations
5
Because the coefficient matrix commutes with itself for all $n$ - you can solve this as a scalar equation then generalize the result. This is not exactly general and a little inelegant, but I can't seem to get Mathematica to do matrix recurrance relations natively. It would be nice if this were supported for general operators.
Anyway we start by noting that
$...
1
Maybe you want FoldList.
rmd = {5.12821, 5.34759, 5.58659, 5.84795, 6.13497, 6.45161, 6.75676, 7.09220, 7.46269}/100.
ints = Prepend[ConstantArray[0.03, Length[rmd] - 1], 0]
bals = FoldList[#1*#2 &, 100, 1 + ints - rmd]
dists = Most[bals]*rmd
2
Here's a rather general recipe that works for a surprisingly large class of problems.
First, define $g_n$ as a recursion (directly, without using RecurrenceTable):
g[0] = x^10;
g[n_] := D[g[n - 1], x]
Compute a few terms:
terms = Table[{n, g[n]}, {n, 0, 10}]
(* {{0, x^10}, {1, 10 x^9}, {2, 90 x^8}, {3, 720 x^7},
{4, 5040 x^6}, {5, 30240 x^5}, {6, ...
6
Given the recurrence
$$
g_n(t) = g'_{n-1}(t),\ \ \ g_0(t) = t^{10}
$$
After applying the Laplace transform we have the transformed sequence
$$
G_n(s) = s G_{n-1}(s),\ \ \ G_0(s) = \frac{10!}{s^{11}}
$$
so
solG = RSolve[{G[n]== s G[n-1], G[0] == 10!/s^11}, G, n][[1]];
Gs = G[n] /. solG;
gt = InverseLaplaceTransform[Gs, s, t]
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