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For the following, solve each of the following recurrence relations by giving explicit formula for $a_n$ and calculate $a_9$.
$a_n = 10 a_{n-1}, a_0 = 3; $
$a_n = -a_{n-1}, a_0 = 5;$
$a_n = 3 a_{n-1} - 1, a_0 = 10; $
$a_n = 4-2 a_{n-1}, a_0 = 0; $
$a_n = 8 a_{n-1} - 15a_{n-2}, a_0 = 1, a_1 = 4; $
$a_n = a_{n-1} +6a_{n-2}, a_0 = 4, a_1 = 4;$

Im trying to self teach my self discrete mathematics and I have just reached recurrence relations but im struggling with it. I was able to do others but im stuck. An clear explanation would be appreciated. There is over 100 exercises in my book so I want to be able to do the other ones my self.

For 3, I got $a_n=(19/2) (3)^n+1/2.$ $For a_9=186989$.
I believe this is correct. But im still struggling to solve for the other ones. Help would be appreciated.

For 1 and 2, I got $3(10)^n$ and $5(-1)^n$ respectively.

user65422
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  • To have more than one character in subscript, use braces {}. I've done this with a_{n-1} in your first line - I'll let you tackle the rest. – 2'5 9'2 Apr 05 '13 at 04:00
  • The first two might become clear if you just write out several terms. Have you tried that? – 2'5 9'2 Apr 05 '13 at 04:01
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    What book are you using? Knowing that would give me a better idea of which techniques that you’re trying to learn at this point. – Brian M. Scott Apr 05 '13 at 04:02
  • For the third and fourth, try substituting $a_n=b_n+c$. For the right values of $c$, they reduce to the form that you see in the first two. – 2'5 9'2 Apr 05 '13 at 04:03
  • For the fifth and sixth there are standard approaches, but they are not something I would have come up with on my own. Someone or some book taught me. – 2'5 9'2 Apr 05 '13 at 04:04
  • For 3, I got $a_n=19/2 (3)^n+1/2. For a_9=186989$. I believe this is correct. But im still struggling to solve for the other ones. Help would be appreciated. – user65422 Apr 05 '13 at 04:17
  • There is a common systematic approach for all these: linear recursive sequences (caracteristic equation for the homogeneous case, undetermined coeffcients for a particular solution in the non homogeneous case). But of course, in easier cases like 1. or 2., it can be done by a simple observation of the pattern of the first terms. – Julien Apr 05 '13 at 04:19
  • A related problem. – Mhenni Benghorbal Apr 05 '13 at 04:51
  • For 1 and 2, I got $3(10)^n$ and $5(-1)^n$ respectively. Is it right? – user65422 Apr 05 '13 at 04:54
  • The whole idea of solving this kind of recurrence equations is to assume the solution as $a_n=r^n$ and substitute back and then solve for $r$. – Mhenni Benghorbal Apr 05 '13 at 05:04
  • @user65422: It is right. – Mhenni Benghorbal Apr 05 '13 at 05:05
  • For 4, I got $2+2(2)^n$ but im very unsure of it. Is there any reference to or example of 5 and 6 you might know of? Those are frustrating me because I can't come up with the solution. – user65422 Apr 05 '13 at 05:10
  • For $(4)$ the solution is $a_n= -\frac{4}{3}, \left( -2 \right)^{n}+\frac{4}{3} $. – Mhenni Benghorbal Apr 05 '13 at 11:33

1 Answers1

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One way to solve the fourth problem is to ‘unwind’ the recurrence:

$$\begin{align*} a_n&=4-2a_{n-1}\\ &=4-2(4-2a_{n-2})\\ &=4-2\cdot4+2^2a_{n-2}\\ &=4-2\cdot4+2^2(4-2a_{n-3})\\ &=4-2\cdot4+2^2\cdot4-2^3a_{n-3}\\ &=4-2\cdot4+2^2\cdot4-2^3(4-2a_{n-4})\\ &=4-2\cdot4+2^2\cdot4-2^3\cdot4+2^4a_{n-4}\\ &\;\vdots\\ &=4-2\cdot4+2^2\cdot4-\ldots+(-1)^{k-1}2^{k-1}\cdot4+(-1)^k2^ka_{n-k}\\ &\;\vdots\\ &=4-2\cdot4+2^2\cdot4-\ldots+(-1)^{n-1}2^{n-1}\cdot4+(-1)^n2^na_0\\ &=4\sum_{k=0}^{n-1}(-1)^k2^k\\ &=4\sum_{k=0}^{n-1}(-2)^k\\ &=4\cdot\frac{(-2)^n-1}{-2-1}\\ &=-\frac43\left((-2)^n-1\right)\\ &=\frac43\left(1-(-2)^n\right) \end{align*}$$

I imagine starting with a large enough $n$ so that I can work back through several steps of the recurrence, and I try to find a pattern. Of course I might have guessed wrong, so at this point I really ought to prove by induction that if I define

$$b_n=\frac43\left(1-(-2)^n\right)\;,$$

then $b_n=a_n$ for all $n\ge 0$. Before doing that, I might make a quick check: this closed formula yields $a_0=0,a_1=4,a_2=-4$, and $a_3=12$, which are easily verified against the original recurrence, so it probably is right.

Another way to solve a first-order linear recurrence with a constant forcing term (the $+4$ in this recurrence) is to make a change of variable. Let $b_n=a_n+d$ for some as yet unknown constant $d$; we’ll choose $d$ so that $b_n$ satisfies a recurrence like the ones in your first two problems. Now $a_n=b_n-d$ (and of course $a_{n-1}=b_{n-1}-d$), which I substitute into the original recurrence to get

$$b_n-d=4-2(b_{n-1}-d)=4-2b_{n-1}+2d\;,$$

or $b_n=-2b_{n-1}+4+3d$. If I let $d=-\frac43$, this becomes $b_n=-2b_{n-1}$, which has the easy closed form solution $b_n=(-2)^nb_0$. And $b_0=a_0+d=-\frac43$, so $b_n=-\frac43(-2)^n$ for all $n\ge 0$. Finally, we reverse the substitution to get

$$a_n=b_n-d=-\frac43(-2)^n-\left(-\frac43\right)=\frac43\left(1-(-2)^n\right)\;.$$

More general techniques will also handle problems like this, but sometimes limited, elementary techniques are quicker.

$a_n = 8 a_{n-1} - 15a_{n-2}, a_0 = 1, a_1 = 4$

I’ll finish up with your fifth problem. The technique most often taught at the elementary level uses a so-called auxiliary equation; in principle this technique works whenever you have a linear homogeneous recurrence with constant coefficients.

To get the auxiliary equation, replace $a_n$ by $x^n$ (and similarly $a_{n-1}$ by $x^{n-1}$ and so on) to get $x^n=8x^{n-1}-15x^{n-2}$. Now divide through by the lowest power of $x$ appearing in the equation to get $x^2=8x-15$. Solve the equation: $0=x^2-8x+15=(x-3)(x-5)$, so the solutions are $r=3$ and $s=5$. This means that the general solution of the recurrence

$$x_n=8x_{n-1}-15x_{n-2}$$

is $$x_n=\alpha 3^n+\beta 5^n\;,\tag{1}$$ where the specific values of $\alpha$ and $\beta$ depend on $x_0$ and $x_1$.

Digression: In general, if the auxiliary equation has distinct solutions $r_1,\dots,r_m$, the general solution will be $x_n=\alpha_1 r_1^n+\ldots+\alpha_m r_m^n$, where the coefficients $\alpha_1,\dots,\alpha_m$ depend on the initial values $x_0,\dots,x_{m-1}$. If it has repeated roots, the general solution is a little different. For instance, if the auxiliary equation had been $x^2-4x+4=0$, the roots would have been $r=s=2$. In that case the general solution would have been $x_n=\alpha 2^n+\beta n2^n$, not $\alpha 2^n+\beta 2^n$, which, after all, is no more general than the single term $\alpha 2^n$, since the two terms can be combined. More generally, if a root $r$ appears $k$ times, it produces terms $$\alpha_1 r^n+\alpha_2 nr^n+\ldots+\alpha_k n^{k-1}r^n$$ in the general solution.

All that remains is to determine $\alpha$ and $\beta$ to get the right initial values. To do this, substitute $t=0$ and $t=1$ into $(1)$:

$$\left\{\begin{align*} &1=a_0=\alpha 3^0+\beta 5^0=\alpha+\beta\\ &4=a_1=\alpha 3^1+\beta 5^1=3\alpha+5\beta \end{align*}\right.$$

Solve this system by whatever means you prefer to find that $\alpha=\beta=\frac12$, so the solution is

$$a_n=\frac12\left(3^n+5^n\right)\;.$$

As a quick sanity check, this does yield the correct values $a_0=1,a_1=4,a_2=17$, and $a_3=76$.

You can now try this technique on the sixth problem.

Brian M. Scott
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