MHB Skyflow3r's question at Yahoo Answers regarding retirement fund

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The discussion focuses on calculating the monthly savings required to accumulate a retirement nest egg using compounded interest at various rates (3%, 6%, 9%, and 12%). Key variables include current age, retirement age, desired nest egg size, and the annual interest rate. A difference equation is established to model the scenario, leading to a solution that expresses the relationship between monthly savings and the accumulated amount over time. The final formula derived for monthly savings is S = (r * E_n) / (12 * ((1 + r/12)^n - 1)), where n represents the total number of months until retirement. This mathematical approach provides a structured way to determine necessary savings for retirement planning.
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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.
 
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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 $$n=12(R-A)$$.
 
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