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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?

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  • $\begingroup$ Let $m=nr$. Then allow $n\rightarrow\infty$. Then observe that $r=k$ $\endgroup$ – CogitoErgoCogitoSum Jul 18 at 19:13
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If I understand the notation in your question, I see a couple of items that seem they should be addressed:

  1. The annual interest rate is $8.12$% which is $r=0.0812$, not $r=0.812$. Also, usually when interest rates are given, they generally refer to "annual" or "yearly" rates.

  2. In the future value calculation, you don't need to multiply $0.0812$ by $12$, because this is already the annual interest rate.

With the above two modifications, one has:

$$ A=8500\left(1+\frac{0.0812}{12}\right)^{12 \cdot 20}=42888.18 $$

I believe to compute the "simple interest" values, one uses the formula:

$$ A_{simple}=8500\left(1+0.0812 \cdot 20\right)=22304 $$

More details here: https://en.wikipedia.org/wiki/Compound_interest#Calculation

I hope this helps.

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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.

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There are some errors in your calculation. First, the value of $k$ is $0.0812$, not $0.812$. Plug this into the exponential growth formula to get $$P(240)=8500\cdot e^{0.0812\cdot 20}\approx 43123,$$ a more reasonable value than $3.67\times 10^{88}$. Second, you substituted into the future value formula incorrectly. Using $r=.0812$ you should get $$ A=P_0\left(1+\frac rm\right)^{mt}=8500\cdot\left(1 + \frac {.0812}{12}\right)^{12 \cdot 20}\approx 42888. $$ Note that these values are close but not exactly the same, because the exponential growth formula $e^{rt}$ is only an approximation to the future value formula $(1+\frac rm)^{mt}$.

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  • $\begingroup$ How did you calculate and get ≈43123 in the first answer? Because I'm calculating again and again but it results ≈ 2.47144x10^12? $\endgroup$ – Nay Sie Jul 18 at 10:25
  • $\begingroup$ @NaySie I fixed a typo. The exponent is $kt$ with $k=.0812$ and $t=20$ in years. $\endgroup$ – grand_chat Jul 18 at 22:20

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