Recursive Def/Compounding Interest question

[rought]

Homework Statement

When Mr. Howe retires at age 65 he expects to have a retirement account worth about $400,000. One month after he retires, and every month thereafter, he intends to withdraw $4000 from the account. The balance will be invested at 9% annual interest compounded monthly.

a) Let A_n represent the amount in the account and n months after Mr. Howe's retirement. Give a recursive definition for A_n

b.) When will there be no money left in the bank account?I know that the recursive formula is something like t_n-1 but that's all I got =/...

 

[tiny-tim]

Hi rought!

a) Let A_n represent the amount in the account and n months after Mr. Howe's retirement. Give a recursive definition for A_n
…
I know that the recursive formula is something like t_n-1 but that's all I got =/...

A recursive formula gives t_n as a function of t_n-1 (and sometimes also t_n-2 etc)

In this case, A_n = … ?

 

[HallsofIvy]

A recursive equation can be written as either [/itex]t_n[/itex]= some function of $t_{n-1}$ or as $t_{n+1}$= some function of $t_n$. Either this amount is some function of last month's or next month's amount is some function of this month's.

Since you mention $t_{n-1}$, let's write it that way. This month's amount, $A_n$ is, first, last month's, $A_{n-1}$, minus any money taken out (how much money does he take out each month?) plus any money put in (interest earned. How much interest does the money earn each month?).

 

[rought]

ah ok...

so here's the equation that i got: A_n=A_n-1(1+.09/12)-4000

Is this correct? I'm not sure how to do part b now though... =/

 

[HallsofIvy]

Okay, $A_n= 1.0075A_{n-1}- 4000$

A standard method for something like $A_n= rA_{n-1}$ is to try something like $A_n= n^x$ for some number. If that were true, then $A_{n-1}= (n^x)^r= n^{rx}$ and $A_n= 1.0075A_{n-1}$ becomes $n^r= 1.0075 (n^{r-1})$. Dividing both sides by $n^r$, $1= 1.0075r^{-1}$ so r= 1.0075. In fact, if we were to try $A_n= C(1.0075)^n$, for C any constant, we would have $A_n= C(1.0075)^n= 1.0075A_{n-1}= 1.0075C(1.0075)^{n-1}= C(1.0075)^n$ is true for all n because the "C"s cancel. $C(1.0075)^n$ is the general solution to the equation $A_n= 1.0075A_n$.

That's ignoring the "-4000" part but since that number is a constant, what if we try $A_n= A$, a constant? Now $A_n= 1.0075A_{n-1}- 4000$ becomes A= 1.0075A- 4000 or -.0075A= 4000.

Now the "theory" part: If $A_n$ is the general solution to the "homogeneous" equation and A is a single solution to the entire equation, then $A_n+ A$ is the general solution to the entire equation.

You should now be able to write out the general solution, use the fact that $A_1= 400000$ to find C and then determine when $A_n= 0$. (You may find that it is never 0. Just find when it is less than 1.)

 

[tiny-tim]

Hi rought!

ah ok...

so here's the equation that i got: A_n=A_n-1(1+.09/12)-4000

Is this correct? I'm not sure how to do part b now though... =/

Yes, that's correct, except that you must also specify the initial condition …

A_n=A_n-1(1.0075) - 4000 and A₀ = 400,000.

You really need to read up about recurrence relations (for example, in the PF Library) to find the general way of solving this.

To get you started, can you see what the solution would be for the simpler:

A_n=A_n-1(1.0075) and A₀ = 400,000?

 

[rought]

ok so I did part b here's what I got

A(n) = 400,000 * 1.0075^n - 4000 * (1.0075^n - 1) / .0075 = 0
400,000 * 1.0075^n = 4000 * (1.0075^n - 1) / .0075


400,000 = 4000 * (1 - 1.0075^-n) / .0075

100 = (1 - 1.0075^-n) / .0075
1 - 1.0075^-n = .75
1.0075^-n = .25
1.0075^n = 4
n * ln(1.0075) = ln(4)
n = ln(4) / ln(1.0075)
n = 185.5315 months

does this seem right?