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I am trying to find a closed-form representation of the following recursively defined sequence

$$π‘Ž_0 = 0, π‘Ž_1 = 0, π‘Ž_2 = βˆ’2, π‘Ž_3 = 0$$ $$π‘Ž_{𝑛+4} = βˆ’2π‘Ž_{𝑛+2} βˆ’ π‘Ž_n$$

I've been working with a lot of homework problems similar to this, but most of the time they are setup in the opposite way, meaning I generally have to find $a_0$ , $a_1$, $a_2$ etc. I am looking for a direction or hint to go in, even a website that may be useful for learning this type of problem would be greatly appreciated. Unfortunately, I am in an online class where it is difficult to get outside help.

I need to go to sleep but will update this in the morning after I attempt the problem. Hopefully I figure it out before I even log back onto my computer!

Thank you all in advance, I've learned so much from using this website in the past week.

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    $\begingroup$ Have a look at the "characteristic polynomial" method. en.wikipedia.org/wiki/…. You will face two little technical difficulties: roots are imaginary and double. For an example of how the method works, you may look at en.wikipedia.org/wiki/Fibonacci_number#Closed-form_expression. $\endgroup$ – user65203 Sep 26 '14 at 10:48
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    $\begingroup$ This is less general than what @Yves suggests, but in this special case, it just might work: calculate $a_4$, $a_5$, and so on, until you see a pattern; write down a closed form formula for this pattern; then prove by induction that this pattern is for real. $\endgroup$ – Gerry Myerson Sep 26 '14 at 10:54
  • $\begingroup$ I spent half of the day today (after a physics exam, which went very well) watching lectures on this material for the question. This is the only question I cannot get and I really appreciate the answer/response from you Macavity @Macavity but I do not understand it. I am at a loss. Never taking 4 accelerated math courses at the same time again. $\endgroup$ – mar10 Sep 27 '14 at 6:09
  • $\begingroup$ The question you just closed was a duplicate, which is not nice, and also a clear-cut case of plagiarism, which is far more serious. $\endgroup$ – Did Sep 27 '14 at 6:26
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You may notice that the recurrence only connects terms with the same parity, so we can split this into two problems which reduce the order of the recursion. Now you could solve the quadratic characteristic equation, or perhaps simply note the pattern as suggested and use induction to prove.

Let $e_n = a_{2n}$ and $o_n = a_{2n+1}$. Then both $e_n$ and $o_n$ satisfy the recurrence $f_{n+2} +2f_{n+1}+f_{n} = 0$, we just have initial values different.

Now $o_0=o_1=0 \implies o_n = 0$ for all $n$. Similarly the pattern $e_n=(-1)^n(2n)$ emerges quickly, and can be proven by induction easily.

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    $\begingroup$ It seems that the OP didn't try very hard... $\endgroup$ – user65203 Sep 26 '14 at 12:14
  • $\begingroup$ @YvesDaoust True. A little effort would have discovered the pattern, writing the closed form perhaps a bit tricky, though proving it should have been easy. $\endgroup$ – Macavity Sep 26 '14 at 12:19
  • $\begingroup$ The GF is also easily written out as $\dfrac{-2 x^2}{(1 + x^2)^2}$, which is even (all odd coefficients vanish) and is readily expanded to get the terms found above. Nothing surprising. $\endgroup$ – Semiclassical Sep 26 '14 at 13:33
  • $\begingroup$ I spent half of the day today (after a physics exam, which went very well) watching lectures on this material for the question. This is the only question I cannot get and I really appreciate the answer/response from you Macavity @Macavity but I do not understand it. I am at a loss. $\endgroup$ – mar10 Sep 27 '14 at 6:03
  • $\begingroup$ Work out the recurrence for the first ten numbers, do you see a pattern? To prove the pattern, induction is enough. Let me know exactly what is not understood $\endgroup$ – Macavity Sep 27 '14 at 6:55

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