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Let $\lambda \in \mathbb{R}$. Is there any way I could solve this recurrence? $$ a_k=-\dfrac{\lambda^2 a_{k-4}}{k(k+1)} $$ where $$ a_0\in\mathbb{R} \quad\quad a_1\in\mathbb{R} \quad\quad a_2=0 \quad\quad a_3=0 $$

LeviathanTheEsper
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  • What is $i$ in the denominator? Is that $\sqrt{-1}$ or the term index? I.e. do you want to solve $$a_k = -\frac{\lambda^2}{k(k+1)}a_{k-4}$$? – uranix May 13 '15 at 15:30
  • Ah, misread. Saw $j$ instead of $i$ in the denominator for no reason. Deleted my answer – Sake May 13 '15 at 15:43
  • $i$ is the index of the sequence. It could be as you say with the $k$, uranix.

    Paul, I was thinking I had forgotten something.

     – LeviathanTheEsper May 13 '15 at 15:53

1 Answers1

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Your sequence actually splits into four distict sequences. Let $i = 4k + l$ where $l = 0, 1, 2, 3$. $$ a_{i} = -\frac{\lambda^2}{i(i+1)} a_{i-4}\\ a_{4k+l} = -\frac{\lambda^2}{(4k+l)(4k+l+1)} a_{4(k-1)+l} = \\ = \frac{\lambda^4}{(4k+l-4)(4k+l-3)(4k+l)(4k+l+1)} a_{4(k-2)+l} = \dots\\\dots = \frac{(-\lambda^2)^k}{\prod_{m=1}^k(4m+l)\cdot\prod_{m=1}^k(4m+1)} a_l = \frac{(-\lambda^2)^k}{16^k\prod_{m=1}^k\left(m+\frac{l}{4}\right)\cdot\prod_{m=1}^k\left(m+\frac{l+1}{4}\right)} a_l = \\ = \frac{(-\lambda^2)^k}{16^k\left(k+\frac{l}{4}\right)_k\cdot\left(k+\frac{l+1}{4}\right)_k} a_l $$ where $(a)_k = a(a-1)\dots(a-k+1)$ is Pochhammer symbol.

uranix
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