1. I need some help with this task. My theory book only shows examples of how to solve sequences in the form :

𝑎𝑘 = A * 𝑎(𝑘−1) − B * 𝑎(𝑘−2).

But I've no idea how to solve this task because of the alternating term. I've included the Answer (called "Svar") to the task.  Reply With Quote

2.

3. Originally Posted by goohu I need some help with this task. My theory book only shows examples of how to solve sequences in the form :

𝑎𝑘 = A * 𝑎(𝑘−1) − B * 𝑎(𝑘−2).

But I've no idea how to solve this task because of the alternating term. I've included the Answer (called "Svar") to the task.
One way to do this would be to replace $a_k$ by $b_k = a_k + c(-1)^k$ (where $c$ is a constant to be chosen later). Then $a_k = b_k - c(-1)^k$, and the recurrence equation for $a_k$ becomes $$b_k - c(-1)^k = 3(b_{k-1} - c(-1)^{k-1}) - (b_{k-2} - c(-1)^{k-2}) - 2(-1)^k,$$ $$b_k = 3b_{k-1} - b_{k-2} + (-1)^k(c + 3c + c - 2).$$ Now choose $c$ so that $5c-2=0$ (so $c = \frac25$). That eliminates the awkward $(-1)^k$ term from the $b_k$ equation, which you should now be able to solve. Having found the answer for $b_k$, you then have $a_k = b_k - \frac25(-1)^k$.  Reply With Quote

Thanks, that was a pretty solution! However the calculations got a bit messy while solving the characteristic equation for bk by hand so I went ahead and used a web calculator for it.

I'm going to give it another shot tomorrow solving it by hand.

We are not allowed to use a pocket calculator at the exam plus you lose a lot of credits if you go wrong somewhere in the calculations. That makes me a really angry student.

Edit: Solved the problem now! thanks again for the elegant solution.  Reply With Quote

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