New answers tagged difference-equations
4
A slick reformulation of bbgodfrey's answer is to recognize that the required computation is equivalent to evaluating an appropriate matrix power:
With[{n = 5},
MatrixPower[{{-m, 1, 0}, {0, 0, 1}, {1, 0, 0}}, n - 2, {1, 0, 0}]] // Simplify
{1 - m^3, -m, m^2}
With[{n = 25},
MatrixPower[{{-m, 1, 0}, {0, 0, 1}, {1, 0, 0}}, n - 2, {1, 0, 0}]] // ...
answered May 26 at 12:17
5
Since you did not provide numerical values, I made some up.
Basically, what you could do is run RecurrenceTable on the 3 equations starting from some initial conditions, then use Graphics3D to plot the trajectory.
ClearAll["Global`*"];
α = 1;
β = 2;
γ = 3;
δ = 4;
ζ = 5;
η = 6;
μ = 7;
ε = 8;
υ = 9;
ρ = 10;
σ = 11;
ω = 12;
eq1 = x[n + 1] == ((α x[n] - β x[n]...
1
One more possibility is to recast your coupled difference equations as repeated matrix-vector multiplication:
$$\begin{pmatrix}p(n+1)\\q(n+1)\\p(n)\\q(n)\end{pmatrix}=\begin{pmatrix}\tfrac{2\beta}{n+1}&-\tfrac{\alpha}{n+1}&\tfrac2{n+1}&0\\\tfrac{\alpha}{n}&\tfrac{2\beta}{n}&0&\tfrac2{n}\\1&0&0&0\\0&1&0&0\end{...
answered May 17 at 13:35
0
One can exploit the equivalence of evaluating three-term recurrence relations with repeated multiplication of $2\times 2$ matrices for this task. Just like in ubpdqn's solution, I use RandomChoice[] to generate a bunch of $\pm1$ multipliers all at once:
BlockRandom[SeedRandom[42, Method -> "MersenneTwister"]; (* for reproducibility *)
With[{n ...
answered May 16 at 14:44
2
I wrote the recursion directly by solving for x[k] in terms of x[k-1]:
Clear[x];
x[k_] := x[k] = (a + k + k^2)/(b - x[k - 1])
x[1] = c;
This clearly gives a "continued fraction-like" answer:
x[#] & /@ Range[10]
Observing that the coefficients in each stack are the same:
coef={6, 12, 20, 30, 42, 56, 72, 90, 110}
I tried:
FindSequenceFunction[{6, ...
4
Clear["Global`*"]
Format[Q[k_]] := Subscript[Q, k];
Format[Q0] = Subscript[Q, 0];
Include the initial condition in RSolve
sol = (RSolve[
{Q[k] == Q[k - 1] + \[Alpha] (Subscript[r, k] - Q[k - 1]), Q[0] == Q0},
Q[k], k][[1]] // Simplify) /. K[1] -> i
Translate the index of summation
sol2 = ((sol /. Sum -> Inactive[Sum]) /.
Inactive[...
6
You are missing the initial condition Q[0] == Q0.
Q[k] /. RSolve[{Q[k] == Q[k - 1] + α (Subscript[r, k] - Q[k - 1]), Q[0] == Q0},
Q[k], k][[1]] // FullSimplify
This is equivalent to
$$(1-\alpha )^k Q0+\sum _{K[1]=1}^k \alpha\ (1-\alpha )^{k-K[1]}\ r_{K[1]}$$
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