Future value formula is:
$A=P \cdot (1+\frac{r}{m})^{m \cdot t}$
where,
- $A$ is resulting amount
- $r$ is annual interest
- $P$ is present value
- $n$ is number of compound periods per year
- $t$ is time (in years)
And, exponential growth function is:
$P(t) = P_0 \cdot e^{k \cdot t}$
The question is:
A retirement account is opened with an initial deposit of $8,500 and earns 8.12% interest compounded monthly. What will the account be worth in 20 years? What if the deposit was calculated using simple interest? Could you see the situation in a graph? From what point one is better than the other?
So to calculate the account worth in 20 years with exponential growth formula:
$P_0$ is $8,500$ and $k$ is $0.812$, months in 20 years is $P(240)$ and so:
for the account worth in 20 years is:
$P(240)=8500 \cdot e^{0.812 \cdot 240} = 3.67052\dots E88$
After calculating with future value formula, the answer is different:
$A = 8500 \cdot (1+\frac{0.812 \cdot 12}{12})^{12 \cdot 20} = 7.71588\dots E65 $
I see different values when I calculate with exponential growth functions and future value formula.
How to achieve this calculation correctly with exponential growth function? Is it possible?
We can use the follwoing approximation. For large $m$ we have $$\left(1+\frac{x}m \right)^{n\cdot m}\approx e^{x\cdot n}$$
With $x=0.0812, m=12$ and $n=20$ the terms are
$$8500\cdot \left(1+\frac{0.0812}{12} \right)^{12 \cdot 20}=42,888.18...$$
$$8500\cdot e^{0.0812\cdot 20}=43,123.4...$$
So the approximation in this case is not so good since $m$ is not large enough. But it goes in the right direction. The larger $m$ is, the closer are the results.