New answers tagged

7

It turns out the Olivier function is already built-in, but in a disguised form: With[{n = 3, a = 2}, Table[j! SeriesCoefficient[1/MittagLefflerE[n, x^n]^a, {x, 0, j}], {j, 0, 40}]] {1, 0, 0, -2, 0, 0, 58, 0, 0, -6218, 0, 0, 1630330, 0, 0, -847053482, 0, 0, 766492673914, 0, 0, -1106653345942538, 0, 0, 2392407356983116538, 0, 0, -...


2

Whenever the current computation depends upon the result of the previous step, think FoldList. FoldList is a fast and efficient function to use for those cases. Let's look at this for three steps where the inputs are symbolic (tip: don't use upper case symbols, you may end up clashing with system symbols). vt[ea_, v0_, cm_, b_] := ea + (v0 - cm)*b ea = {...


1

RecurrenceTable[{vt[n + 1] == EA + (vt[n] - Cm)*B, vt[1] == 3500}, vt, {n, 1, 10}]


2

If you are interested in increasing performance I would recommend using FindMaximum. If you go through five steps you will discover at the end the original account balance (a0) is related to the five withdrawals via: a0 == (w5 + w4 z + w3 z^2 + w2 z^3 + w1 z^4)/z^4 where w1 through w5 are the withdrawal values and z is the interest rate. It is desired ...


2

f1[n_] = Sum[Fibonacci[i]*(-1)^i, {i, 0, n}]; If you know or suspect the general form of the model model = (-1)^n Fibonacci[b*n + c] + d; data = Table[f1[n], {n, 10}] // Simplify; f2[n_] = model /. ( NonlinearModelFit[data, model, {b, c, d}, n][ "BestFitParameters"] /. x_?NumericQ /; (Abs[x - Round[x]] < 10^-10) :> Round[x]) (* -1 + ...


4

There is probably no package or algorithm that can give you answer in all cases. However, I would like to suggest a semi-automatic solution. I will illustrate it with few examples. Notice, I do not know solutions in advance. Example 1 t[1] = Table[Sum[Fibonacci[3 i], {i, 0, k}], {k, 0, 10}] *({0, 2, 10, 44, 188, 798, 3382, 14328, 60696, 257114, 1089154}*) ...



Top 50 recent answers are included