We have the following formula when working with interest rates:
$$FV\:=\:PV\cdot\left(1+\frac r{100\,k}\right)^{kn}$$
where $FV$ = future value, $PV$ = present value, $n$ is the number of years, $k$ is the number of compounding periods a year, and $r$ in % is the nominal annual rate of interest.
Now, I could, technically, just apply this formula blindly. But I prefer to understand things and, if nothing else, feel that this will help me understand it better. Therefore:
Could anyone please explain intuitively where this formula comes from? Why are we dividing $r$ by $k$ and then raising to the power of $k$? Why not just multiply $PV$ by $(1+\frac r{100})^{n}$?
I suppose this has something to do with the fact that $r$ is a nominal annual interest rate, but then why not just state the 'correct' annual interest rate to begin with… an example of this is in the first answer below when we discuss a 20% nominal interest rate.
Help would be much appreciated!
Best Answer
Well to start the 100 is just a way to convert the percentage value and isn't really necessary as percentages are typically understood to be decimals anyway.
So now we're left with $$FV = PV\left(1+\frac{r}{k}\right)^{kn}$$
The intuition here is quite simple, you're just considering '$k$' extra periods per year at '$\frac{1}{k}$'th of the rate. for example instead of an interest rate of $10\%$ per annum for $10$ years we can think of it as $5\%$ per 'half year' with twice as many half years ($20$ half years).
So now we're back down to trying to understand:
$$FV = PV\left(1+r\right)^n$$
Which is simply just stating that the future value of an investment (or loan or whatever else you're compounding) is its present value, iteratively multiplied (n times) by some factor $(1+r)$. And this is just the way we increase a value by a percentage. For example if I wanted to increase $100$ by $20\%$ I would say $100\cdot\left(1+20\%\right)$ which is equal to $120$. If I wanted it to grow twice, I would multiply it again so:
$$100\cdot\left(1+20\%\right)\cdot\left(1+20\%\right)$$
or, equivalently: $100\cdot\left(1+20\%\right)^2$
If I wanted to grow it n times, I would say:
$$100\cdot\left(1+20\%\right)\cdot\left(1+20\%\right)\cdot...\cdot\left(1+20\%\right)\qquad\text{'n' times}$$
or, equivalently: $100\cdot\left(1+20\%\right)^n$
Bringing this idea back to the original formula you shared, if I changed the context to compounding $\$100$ at $20\%$ per annum for $n$ years, but I wanted to compound it every $6$ months ($2$ times a year) I would essentially split my $20\%$ rate in half and compound it twice as many times:
$$100\cdot\left(1+\frac{20\%}{2}\right)^{2n}$$
Where here $k$ from the original formula is $2$.
It is important to understand that this is different from taking the yearly rate and raising it to the power of the number of years because of the nature of compound interest. Every time an investment is compounded, it's growing by it's current value as opposed to growing by it's initial value (like in simple interest) So in the scenario above, compounding it twice a year means that it can grow for the first half of the year and then the next half grows from whatever value the investment had at the half way point (which for positive interest rates will be more than the initial value) so it grows faster. The more compounding periods we have, the greater the $FV$ will be if everything else remains constant (to a limit: if you want to learn more about this limit you can research the number $e$).
$\textbf{Important for building intuition}$
If you want to visualise, this hop onto desmos, or any other graphing website/software, and type in: $y=a(1+\frac{r}{x})^{xn}$ into one line, and $y=a(1+r)^n$ on another. You can set values yourself for present value (a), rate (r) and number of periods (n) or you can just add a slider for them and experiment a little bit!
You should be able to see that when $x>1$, i.e. we're compounding more frequently than once per year, the formula with compounding is greater than the formula without.
Just for fun, while you're on Desmos, type in $y=ae^{rn}$ in another line and you'll see that the compounding function approaches $e$ but never surpasses it - that's what I was referencing earlier on when I mentioned a limit.