Recurrence relation - How to solve this recurrence relation

Question

a person invests 1000 at a bank at 4 percent compound interest compounded annually and every year government and bank charges amounting to C are deducted and if An is the value of the investment at the end of 10 years.

Solve this difference equation.

if i)C=0 ii)C=40

$$ A_{10} = 1.04 A_{n-1} - C $$
$$ A_0 = 1000 $$

i) C=0

$$ A_{10} = 1.04^{10} . 1000 = 1480.24 $$

ii) C=40

$$ A_n = 1.04 . A_{n-1} - 40 $$

Particular solution: Put $$ A_n = A_{n-1} = A* $$
$$ A* = 1.04 . A* - 40 $$
$$ A* = 40/0.04 = 1000 $$

General solution of the associated homogenous equation: $$ A_n = 1.04 A_{n-1} $$
$$ a_n = A . 1.04^n $$

General solution of the difference equation: $$ A_n = A_n + A* $$
$$ A_n = A . 1.04^n + 1000 $$
$$ A_0 = A . 1.04^0 + 1000 $$
$$ A = 1000 $$

$$ A_{10} = 1000 . 1.04^{10} + 1000 $$
$$ = 1480.20 + 1000 $$
$$ = 2480.24 $$

Getting 2480.24 as the answers tell me something is wrong. But I cant figure out which part of my working is incorrect.

Since 1000-40 for 10 years I believe the answer should be

$$ A_{10} = 1.04^{10} . 1000 - 40 $$

Please advise.

Answer 1

I propose that we deal with this problem in a purely algebraic fashion, i.e., without talking about specific numbers like 4% or $40\$$ or 10 years. So we have an initial investment $a_0$ on January 1st of year 1, an annual increment factor $1+p$ and an annual charge $c$. After $n$ years have elapsed, i.e., on January 1st of year $n+1$, the amount on the account is $a_n$. The sequence $n\mapsto a_n$ satisfies the recurrence relation $$a_{n+1}=(1+p)a_n - c\qquad(n\geq 0)\ .$$ If it weren't for the $c$ the $a_n$ would increase exponentially. To account for the charges we try a solution of the form $$a_n= a (1+p)^n + b\qquad (n\geq 0)\qquad(*)$$ and hope that we can fix the constants $a$ and $b$ in such a way that all conditions of the problem are fulfilled.

Putting $n=0$ in $(*)$ we must have $a+ b=a_0$ (where $a_0$ is given in advance), and the recurrence relation implies $$a(1+p)^{n+1} + b = (1+p)\bigl(a(1+p)^n + b \bigr) -c$$ from which we draw the condition $b=(1+p) b -c \ $ or $\ b={c\over p}$. It follows that $a=a_0-{c\over p}$ so that we definitively obtain $$a_n=\Bigl(a_0-{c\over p}\Bigr)(1+p)^n + {c\over p}\ .$$

Answer 2

Your work on i.) is correct. For ii.) you could just notice that with $C=40, A_0=1000$, you get $A_1=1000$ and nothing changes. So $A_n=1000$ for all $n$. If $A_0$ were not $1000$, you could imagine breaking $A_0$ into two pieces-$A_*$ and $A_{**}$. The $A*$ piece, we see, doesn't change and pays the annual fee of $C$. The $A_{**}$ piece (which could be negative) is left to grow at $4\%$ per year. So if you started with $A_0=1500$ you would say $A_*=1000, A_{**}=500,$ so $A_n=1000+500\cdot1.04^n$. Have you checked out the Wikipedia article?

Answer 3

In general: We know the recurrence:

$a_{n+1} = r \; a_n $

has the explicit solution

$a_n = r^n a_0$

To solve the non-homogeneous recurrence

$a_{n+1} = r \; a_n + c $

we seek an alternative $b_n = a_n + x$ ($x$ is to be found) such that the above equation is equivalent to

$b_{n+1} = r \; b_n $

Replacing, we must have $c = rx -x$, so $x = c/(r-1)$

And then the solution is

$b_n = r^n \; b_0$

$a_n + x = r^n \; ( a_0 + x)$

$\displaystyle a_n = r^n \; a_0 + c \frac{ r^n-1}{r-1}$