Solving a non-homogenous linear recurrence relation with initial conditions

Question

Given the recurrence relation:

$$ a_n = 1.04a_{n-1} + 100 $$ $$ a_0 = 0 $$

How can I solve this? Here is my attempt at a solution:

1. Solve the non-homogeneous part. Set up a polynomial $p_n$: $$ p_n = D $$ $$ D = 1.04D + 100 $$ $$ -0.04D = 100 $$ $$ \therefore D = -2500 $$

2. Solve the homogeneous part. This part appears to be correct (per my textbook's listed solution):

$$ a_n = 1.04a_{n-1} $$ $$ Characteristic \ polynomial: \ x = 1.04(x)^0 $$ $$ \therefore x = 1.04 $$

3. Add the solutions:

$$ a_n = C_1(1.04)^n - 2500 $$ $$ For \ n=0: a_0 = 0 = C_1(1.04)^0 - 2500 $$ $$ C_1 = 2500 $$ $$ \therefore a_n = 2500(1.04)^n - 2500 $$

Right? But my book says the solution is $a_n = 1250(1.04)^n - 1250$. How can this be? Did I do something incorrectly?

Answer 1

We can solve it with a simple substitution, which really gives an intuition on why your method works.

$$a_n = 1.04a_{n-1} + 100$$

$$-2500=1.04(-2500)+100$$

Subtracting both equations gives,

$$a_n+2500=1.04(a_{n-1}+2500)$$

Let $b_n=a_n+2500$ this gives,

$$b_n=1.04b_{n-1}$$

$$b_n=b_0(1.04^n)=(a_0+2500)(1.04^n)$$

Which means,

$$a_n=(a_0+2500)(1.04^n)-2500$$

Answer 2

Your solution is good. I'll present a slightly different approach.

Let's consider the more general $a_{n+1}=ha_n+k$, $a_0=0$. We have $$ a_{n+2}=ha_{n+1}+k=ha_{n+1}+a_{n+1}-ha_n=(h+1)a_{n+1}-ha_n $$ which is a second order recursion, with initial data $a_0=0$, $a_1=k$.

The characteristic polynomial is $X^2-(h+1)X+h$, whose roots are $h$ and $1$. Thus the general solution, assuming $h\ne1$, is $$ a_{n}=Ah^n+B $$ and we have \begin{cases} A+B=0\\ Ah+B=k \end{cases} hence $B=-A$ and $A(h-1)=k$, so $$ A=\frac{k}{h-1},\quad B=-\frac{k}{h-1} $$

For $h=1.04$ and $B=100$, we have $$ A=\frac{100}{0.04}=2500,\qquad B=-2500 $$

Answer 3

Your solution is correct. You can tell that the book's solution is wrong with a counter-example:

Using the solution given by the book (Letting $n=1$): $$a_1=1250(1.04)^1-1250=50$$ But if we substitute this into your recurrence relation: $$a_1=1.04a_0+100$$ $$50\neq 100$$ And thus, the solution provided by your book is incorrect.

You can prove that the solution you've obtained is correct using Mathematical Induction.

Answer 4

I'd like to present and alternative which relies on reducing the problem into something simpler, by removing the constant in front of $a_n$.

In general assume we want to solve $a_{n + 1} = ba_n + k$. We divide both sides by $b^{n+1}$ so that we get

$$\frac{a_{n + 1}}{b^{n+1}} = \frac{a_n}{b^n} + \frac{k}{b^{n + 1}}$$

This essentially transforms our problem into solving

$$c_{n + 1} = c_n + \frac{k}{b^{n + 1}}$$

where $c_n = \frac{a_n}{b^n}$.

This is pretty straight forward: We move $c_n$ to the left hand side:

$$c_{n + 1} -c_n = \frac{k}{b^{n + 1}}$$

We then sum up both sides from $n = 1$ to $n = N$. On the left, we observe the simplest ever telescoping series and to on the right hand side, we observe a rather simple geometric series.

With this we get a closed expression for $c_{N + 1}$ and can extract $a_{N + 1}$ with a simple mulitplication of $b^{N + 1}$ on both sides.

Hopefully this comes in useful when you forget any formulas pertaining to the solutions or characteristic polynomials or what not.