2
$\begingroup$

Need help solving the following system: $$s_n=s_{n-1}+2t_{n-1}+2$$ $$t_n=-2s_{n-1}+t_{n-1}+n$$ $$n\ge2$$ $$s_1=1$$ $$t_1=2$$

$\endgroup$
  • $\begingroup$ Just input $\ t_n$ into the equation for $\ s_n$ $\endgroup$ – CTSnake May 2 '17 at 14:41
  • $\begingroup$ Write it as a matrix equation. $Q_n=M Q_{n-1} + K_n$ $\endgroup$ – user121049 May 2 '17 at 14:50
  • $\begingroup$ You can solve either by obtaining a recurrence purely in $s$ or purely in $t$ then once that is solved you can use one of your two recurrences to solve for the other. You can start by substituting from one equation into the other, you decide on the most appropriate choice. Please post any of your attempts so that the mathSE community is in a better position to help you. $\endgroup$ – N. Shales May 2 '17 at 14:54
  • $\begingroup$ Z transforms will do the trick too. $\endgroup$ – Paul May 2 '17 at 15:00
0
$\begingroup$

Hint

$$s_n=s_{n-1}+2t_{n-1}+2\quad (1)\\ t_n=-2s_{n-1}+t_{n-1}+n\quad (2)$$

make $-2\cdot(2)+(1)$:

$$s_n-2t_n=5s_{n-1}+2(1-n)\to t_n=\frac{s_n-5s_{n-1}-2(1-n)}{2}$$

replacing at $(1)$:

$$s_n=s_{n-1}+2t_{n-1}+2\to s_n=s_{n-1}+s_{n-1}-5s_{n-2}-2(2-n)+2$$

$$s_n-2s_{n-1}+5s_{n-2}+2(1-n)=0$$

Can you finish?

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.