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Let $r>4$ be a positive integer. I want to solve this difference equation:

$$u_{n+1}-r²(1+r²ⁿ⁺¹)u_{n}+r²r²ⁿ⁺¹u_{n-1}-2r²r²ⁿ⁺¹=0$$ but I have no a good idea to start.

DER
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  • Isn't this a minor variation on your previous question http://math.stackexchange.com/q/1138033 ? – Did Mar 01 '15 at 18:43
  • @Did: Yes, but the method does not works well. – DER Mar 01 '15 at 18:44
  • @Did: No, the two questions are different. – DER Mar 01 '15 at 18:45
  • "Yes, but the method does not works well." ?? Which method? What are you talking about? Why didn't you mention the other question? – Did Mar 01 '15 at 18:45
  • @Did: You can see the usual method of summation to reduce it to a first order equation does not works for this case. – DER Mar 01 '15 at 18:46
  • Where can I see "the usual method" since you do not present it, neither here nor there? – Did Mar 02 '15 at 06:12

1 Answers1

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I expanded your recurrence

$$ u_n = \left(r^2 + r^{2n+1}\right) u_{n-1} - \left( r^{2n+1} \right) u_{n-2} + 2r^{2n+1} $$

using Sympy and found a general solution in terms of $u_0$ and $u_1$.

$$ u_n = u_1\, r^{f(n)} + \sum_{k=0}^{n-2} \left[ \left( 2k+2-u_0 \right) r^3 + u_1 \right] r^{f(n)-g(k)} \\ \; \\ \begin{array} ( &f(n) = n^2 + 2n - 3 \\ &g(k) = k^2 + 4k + 3\\ \end{array} $$

which can be proved by induction. However, unless you can simplify this further, it's not much more computationally useful than the recurrence. As an example, here is the expanded form of $u_{10}$ when $u_0=u_1=0$.

$$ u_{10} = 2r^{117} + 4r^{112} + 6r^{105} + 8r^{96} + 10r^{85} + 12r^{72} + 14r^{57} + 16r^{40} + 18r^{21} $$

Andrew Szymczak
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