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A standard derivation of the number $e$ is as continuous compound interest. For example, Wikipedia says, of an annual 100% rate:

If the interest is credited twice in the year, the interest rate for each 6 months will be 50%

Because 50 = 100 / 2. If it was monthly, we'd divide by 12 instead to get 8.3%. I do accept that, if we follow this reasoning, we get to the number $e$. And I used to follow it just fine, but when I just revisited it, I had a problem.

I don't understand why a competent bank would calculate interest rates in such a way that I'd get more money depending on how finely they happen to divide up the year! Surely the correct calculation for a month is to take $200\%^\frac{1}{12} \approx 1.059$ or 5.9%, in which case the final balance is unchanged. Isn't it incorrect to posit an equivalence of an annual 100% rate and a monthly 8.3% rate? The growth is different: $(1 + 0.083)^{12} \approx 2.6 \neq 2.0 \approx (1 + 0.059)^{12}$.

This story doesn't work for me as a "natural" account of how one would discover the number $e$. Yet I can't deny that the formula

$$ e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n $$

really does contain a $\frac{1}{n}$ and no trace of an $n$th root.

So is there some natural reason why we'd use an "arithmetic" compound interest division over a "geometric" one (other than "because it leads us to the definition of $e$"?) What am I missing?

Clarification: Say the initial deposit should scale by $A = 1 + r$ after a year. Say that this gets paid $n$ times in this period at regular intervals. In the story, the bank divides by n to get a monthly scaling factor of $A' = 1 + r/n$. This is wrong! It means the deposit will grow by more than the promised $A$ at the end of the year, and the bank will give the customer more money than they agreed! The correct way is to take the $n$th root, so $A' = A^{1/n}$. This way, at the end of the year, the final scaling is $(A')^n = A$ as set out at the beginning.

It appears that $e$ is the natural result of continuous compound interest only if you calculate interest in an economically irrational way. Am I correct or do real-world banks calculate interest by division for some sensible reason?

jdjakub
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  • When you invest 100 units of a currency into a bank account that pays you monthly interest $r$ and that $r$ is expressed as percent per year. Then how much money is in your account after two months ? – Kurt G. Oct 27 '22 at 09:55
  • Not sure what you are asking. In practice, deposits are paid interest at regular intervals (just like bonds pay coupons at preset intervals). It would be inconvenient to have continuously changing deposit values. Everyone knows the math works out better if you use continuous compounding but reality is what it is. – lulu Oct 27 '22 at 11:13
  • @KurtG. I don't understand your first sentence. So is r a monthly interest rate or an annual one? Assuming it's monthly, and we have the convention r > 1, after two months I'd have $ 100r^2 $ in the account. – jdjakub Oct 28 '22 at 10:39
  • @lulu My question isn't about discrete vs continuous compounding but about arithmetic diminishment (division; dividing by n) versus geometric diminishment (taking an nth root). If I was a bank and I offered my customer 5% interest annually, but decided to pay it every 6 months, I'd scale their deposit by $\sqrt(1.05) \approx 1.024$ i.e. pay 2.4% interest. This is below the 2.5% I'd get by just dividing 5% by 2. Is that clearer? – jdjakub Oct 28 '22 at 10:43
  • No, it isn't. This is just about how rates are quoted, and you can quote them any way you like. If the bank pays you $x%$ $n$ times a year, the simplest way to quote that as an annual rate is to add up the payments, to get $n\times x$. Of course this ignore the effect of compounding but compounding isn't clear (as you might not be able to reinvest at the same rate). So quoting simple interest makes sense. You could some other way if you wanted, but it doesn't change anything, – lulu Oct 28 '22 at 10:50
  • Thanks. I followed except for "compounding isn't clear (as you might not be able to reinvest at the same rate)" - what does this mean? Why not? – jdjakub Oct 28 '22 at 11:03
  • @jdjakub . I have never seen an interest rate $r$ that is strictly greater than $1$ (which means strictly greater than $100%$). Typical interest rates are around $5%$. They are quoted as *percent per year* (as I wrote). When the rate is paid every month you have in your account $100(1+r/12)$ after the first month and $100(1+r/12)^2$ after the second. – Kurt G. Oct 28 '22 at 13:01
  • What banks typically call the "annual interest rate" is actually the nominal annual interest rate; as such, division by $n$ is consistent with its definition. On the other hand, $n$-rooting is related to the effective annual interest rate. (These two quantities are related by $n.$) – ryang Aug 06 '24 at 12:32

1 Answers1

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The issue is one of semantics and terminology, not mathematics or economics.

When we talk about concepts like "continuous compounding" or even interest rates in general, it is important to be precise about what we mean, because as is so often the case in mathematics, formulas and equations can disambiguate what is not always clearly expressed through language.

For instance, when we think about interest earned on a fund, it is often convenient to conceptualize the amount earned over some fixed period of time, usually a calendar year. So for instance, if I suppose I invest $1000$ that earns $5\%$ interest per year, then at the end of one calendar year I would have $1050$. And if I reinvest this amount at the same rate, then at the end of the second year I would have $1102.50$, and so forth. This is the familiar "compound interest" that we have seen before:

$$A(n) = A(0) (1 + i)^n,$$ where $A(0)$ is the original amount invested, $i$ is the annual rate of interest, and $n$ is the whole number of years the fund is invested.

The more precise terminology for the value $i$, however, is the effective annual rate. This is the amount, irrespective of how the fund might be compounded during a year, by which the fund effectively grows.

But suppose we want to check on the value of our fund at the end of each calendar month, rather than each year. The bank, as you put it, isn't giving us more or less money; the rate of return on our investment is the same whether or not we look at the balance at some intermediate point in time. If we assume that the interest rate is not fluctuating over time, how much can we expect to see at the end of each month? That is to say, what is the effective monthly rate that is equivalent to the effective annual rate? This requires us to find the solution to the equation $$(1 + j)^{12} = 1 + i,$$ where $j$ is the desired effective monthly rate. We get $$j = (1+i)^{1/12} - 1,$$ and for $i = 0.05$, this gives $j \approx 0.00407412$ or about $0.407\%$ per month. So at the end of the first month, you would see a balance of $1004.07$ in the account; at the end of the second, you would have $1008.16$, and so forth, until at the end of the twelfth month you would again have $1050$ exactly.

And we can extend this reasoning for finer time increments; e.g., if we want to check on the balance daily, the effective daily interest rate (assuming 365 days in a year) is about $0.0133681\%$. Again, we are assuming that interest is compounded at a constant rate (whereas in practice, interest rates fluctuate with market conditions, but this is more a topic for economics as opposed to mathematics).

However, effective rates for extremely short time periods are not convenient to report. As you can see, their value tends to $0$ as the time increment also tends to $0$ (as an exercise, what is the effective hourly rate, assuming 365 days in a year and 24 hours in a day?). This is where nominal rates of interest come into play. A nominal rate is defined to be the effective rate multiplied by the number of compounding periods over which the rate is converted; e.g., the nominal monthly rate that is equivalent to an effective annual rate of $5\%$ is $12(0.00407412) \approx 0.0488895$ or about $4.89\%$. We use the symbol $i^{(12)}$ to describe this rate; therefore, $$\left(1 + \frac{i^{(12)}}{12}\right)^{12} = 1 + i$$ and in general, the nominal $m^{th}$-ly rate that is equivalent to an effective annual rate of $i$ is given by $$i^{(m)} = m\left((1 + i)^{1/m} - 1\right).$$ (Note that the use of parentheses in the superscript is so that we do not confuse it for an exponent.) So in the daily case, $i^{(365)} \approx 0.0487934$ and in the hourly case, $i^{(8760)} \approx 0.04879030$. You can see that as the compounding frequency increases, the nominal rate tends towards a nonzero fixed number. In the limit, we get $$i^{(\infty)} = \log(1 + i) \approx 0.048790164169432.$$ This is what we call the force of interest and instead of $i^{(\infty)}$ we typically write $\delta$.

Again, every single rate, whether nominal or effective, monthly or daily or hourly, I have calculated so far in our discussion, is equivalent to the same effective annual rate of $5\%$, in the sense that at the end of the first year, the balance is still $1050$ on an initial investment of $1000$.

Now that we have properly described our notation and terminology, where does $e$ come from? Well, up to this point we have been characterizing the same effective interest rate in different ways by considering the equivalent nominal rate or the equivalent rate over different time periods. In this way, we still end up with the same return--the bank is not giving you any more or less money.

However, what if instead we consider the same numerical value for the nominal interest rate, but compound it over different periods of time? That is to say, for a given fixed nominal rate, say $5\%$, what is the equivalent effective annual interest rate as the conversion/compounding occurs more frequently?

In the case where $i^{(12)} = 0.05$, so we have a nominal monthly rate of $5\%$, the effective annual rate is $$i = \left(1 + \frac{i^{(12)}}{12}\right)^{12} - 1 = \left(1 + \frac{0.05}{12}\right)^{12} - 1 \approx 0.0511619,$$ or about $5.12\%$. If the nominal daily rate is $i^{(365)} = 5\%$, then the effective annual rate is $$i = \left(1 + \frac{i^{(365)}}{365}\right)^{365} - 1 \approx 0.0512675,$$ and so forth. Unlike our earlier discussion, each of these situations results in a different accumulated value of the fund at the end of the year because the effective rate of interest is different: the more frequently the same nominal rate of interest is converted/compounded, the more money is earned. With annual compounding, the amount is $1050$; with monthly compounding, you would have $1051.16$; with daily compounding, you would have $1051.27$. But as you can see, as the compounding frequency increases without bound, there is a limit to the amount of money you will ultimately earn in a year, just like in our earlier discussion about the force of interest: If you compound infinitely frequently, you only get about $1051.271096$. What is this value in relation to $5\%$ and $1000$? It is $$1000 e^{0.05}.$$ This is because in this case, the force of interest is $\delta = 0.05$.

So to emphasize, in the first part of our discussion, the effective rate and hence the annual return is fixed, while the nominal rate is allowed to vary with the compounding frequency, until we arrive at some equivalent force of interest. In the second part of the discussion, the nominal rate and hence the force of interest is fixed, while the effective rate and annual return are allowed to vary.

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