New answers tagged

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]


Top 50 recent answers are included