What exactly does it mean? By continuously compounded it makes me think it is almost like multiplied as time goes on. Could someone also explain what the constant e is and how it originated? Also how do they get the compounded interest formula from the constant e?
Continuously Compounded Interest
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So lets look at the what we mean by compound interest. Let use $r$ as a rate for a year. If we want to define a semiannual compound rate we do the following $$ \left(1+\frac{r}{2}\right)\left(1+\frac{r}{2}\right) = \left(1+\frac{r}{2}\right)^{2} $$ for quarterly we have $$ \left(1+\frac{r}{4}\right)^4 $$ since at the end of the year we have $4$ growths where we have invested the increased value. in general we have for $n$-periods we have $$ \left(1+\frac{r}{n}\right)^n $$ now if we have a continuous growth then we are effectively saying that period becomes infinitesimally small i.e. $n\to\infty$ Mathematically speaking $$ \lim_{n\to\infty}\left(1+\frac{r}{n}\right)^n\to e^r $$ where $\mathrm{e}$ is the exponential constant. A really special function in mathematics is $\mathrm{e}^x$ where the gradient at any point of the graph $\mathrm{e}^x$ is proportional to itself, so ideal for growth formula.
Now we can extend this from one year to many years by using $$ \left(1+\frac{r}{n}\right)^{nt} $$ where we can adjust the index by $nt=k$ we get $$ \left(1+\frac{rt}{k}\right)^{k} $$ if we have $r,t<\infty$ then we have $$ \lim_{k\to\infty}\left(1+\frac{rt}{k}\right)^{k} \to \mathrm{e}^{rt} $$
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The other replies offer great explanations for how the formula is generated and e's purpose, but this is the simple formula you will use for interest that is compounded continuously:
$ A=Pe^{rt} $
Where A is Amount, P is Principal (money down), e is a constant (~2.718281828), r is rate, and t is time.
As far as what compound interest is, think of it like this. Say you invest 500 dollars and at the end of each year, you get half of your initial investment added to your bank account. After 1 year, you have 750 dollars, and after 2 years, you have 1,000 dollars, etc. Simple interest is linear.
If this were interest compounded annually, however, you would have 750 dollars after the first year, but after the second, you add 50% of the 750 dollars (not the principal of 500 dollars like with simple interest) and you now have 1,125 dollars after two years, not 1,000 dollars like with simple interest. If you graph this, you'll soon see that rather than having a linear shape like simple interest, it has the shape of an exponential function.
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When compounding for the amount not annually, not bi-annually, monthly or daily or by the hour or by the second... the amount does not go to $\infty$ but mathematically tends to definition of the exponential function.
$$ A = P e^{r\,t} \leftarrow A = P + P \,r\, t $$ for annual compounding, where we have taken first two terms in an infinite series as an approximation.
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As the frequency of compounding increases, the amount you have after any specified time increases, but not without bound. For any specified time period and interest rate, there are amounts that cannot be exceeded by the final balance no matter how frequently you compound. The smallest of those amounts that cannot be exceeded by discrete-time compounding is the amount earned in continuous compounding.
Let denote the interest per annum (year) as $i$. Therefore the compound interest is $1+i$. Now you can compound twice a year. But you take only the half of the interest rate, because the period between the compoundings are only half as long as before.
$\left(1+\frac{i}{2} \right) ^2$
Then you compound 4 times a year (every three months). Thus you divide i by 4 and exponentiate it by 4.
$\left(1+\frac{i}{4} \right) ^4$
Now imagine you do compound infinitely times in a year. The expression becomes
$$\lim_{n \to \infty}\left(1+\frac{i}{n} \right)^n=e^i $$
This means that the interest per year is effectively $(e^i-1)\cdot 100\% \neq i$
$\color{blue}{\text{numerical example}: i=0.05}$
$(e^{0.05}-1)\cdot 100\% =5.127\%\neq 5\%$
You see, that this tansformation from yearly compounding to continuous compounding is not equivalent.