Skyflow3r's question at Yahoo Answers regarding retirement fund

[MarkFL]

Here is the question:

Calculus Question > compounding interest?

How much is needed to save each month at 3,6,9, and 12% compounded monthly for you to accumulate a nest egg for retirement.

Variables are age, age of retirement, nest egg size and interest rate.

Thanks for any help

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello skyflow3r,

First, let's define the variables:

$A$ = present age.

$R$ = retirement age.

$E_n$ = size of nest egg, or account balance after $n$ months.

$r$ = annual interest rate.

$S$ = the amount saved and deposited monthly.

We may model the given scenario with the following difference equation:

$$E_{n+1}-\left(1+\frac{r}{12} \right)E_{n}=S$$ where $$E_1=S$$

We see that the homogeneous solution is:

$$h_n=c_1\left(1+\frac{r}{12} \right)^n$$

And we seek a particular solution of the form:

$$p_n=k$$

Substituting the particular solution into the difference equation, we obtain:

$$k-\left(1+\frac{r}{12} \right)k=S$$

$$k=-\frac{12S}{r}$$

Thus, by superposition, we find:

$$E_n=h_n+p_n=c_1(1+r)^n-\frac{12S}{r}$$

Using the initial value, we may determine the parameter:

$$E_1=c_1\left(1+\frac{r}{12} \right)^1-\frac{12S}{r}=c_1\left(1+\frac{r}{12} \right)-\frac{12S}{r}=S\,\therefore\,c_1=\frac{12S}{r}$$

And so, we find the solution satisfying all of the given conditions is:

$$E_n=\frac{12S}{r}\left(1+\frac{r}{12} \right)^n-\frac{12S}{r}=\frac{12S}{r}\left(\left(1+\frac{r}{12} \right)^n-1 \right)$$

Now, solving for $S$, we obtain:

$$S=\frac{rE_n}{12\left(\left(1+\frac{r}{12} \right)^n-1 \right)}$$

where $$n=12(R-A)$$.