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Find the recurrence relation for: $a_k = -4_{k-1}-4_{k-2}$ when $a_0=0$ and $a_1=1$

Step 1: $r^k=-4r^{k-1}-4r^{k-2}$

Step 2: $0= r^2+4r+4 = (r+2)^2$

$r_1=r_2=-2$

$a_k=A(r_1)^k +Bk(r_2)^k$ (when roots are equal)

$a_k=A(-2)^k+Bk(-2)^k$

$a_0=0=A(-2)^0+B(0)(-2)^0$

$a_0=0=A+B(0)1 = A=0$

$a_1=1=A(-2)^1+B(1)(-2)^1$

$a_1=1=-2A-2B=-\tfrac{1}{2}=A+B$

$a_1-a_0$:${0=A+0B}\over {-\tfrac{1}{2}=A+B}$=$\tfrac{1}{2}=B$

If $B=\tfrac{1}{2}$ Then $A+B=0$ then $A=-\tfrac{1}{2}$

So: $a_n=-\tfrac{1}{2}(-2)^n+\tfrac{1}{2}(-2)^n$

Does this look correct? I'm new to recurrence relations.

Demosthene
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DavidC85
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1 Answers1

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You started off great (although you might want to learn how to insert an implies symbol as in $\text{this } \implies \text{ that}$ so you are not writing something that looks like $-1 = -\frac12$).

You went wrong after yourcorrect equation $A+B = -\frac12$. You already had correctly shown that $A = 0$ so you should have then said that $B = -\frac12$.

And then the solution would be $$ a_n = -\frac{n}{2} (-2)^n $$

Mark Fischler
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  • Awesome, thanks. I got that and thought it wasnt correct because then half of the initial equation would be missing. Guess I should trust the math. – DavidC85 May 02 '15 at 20:32