# Why is e (Eulers constant) included in the formula for continuous interest?

Lets start with the most basic continuous interest formula and the evaluation of e:

Continuous interest: $\left(1+\frac{r}{n}\right)^n$

Evaluation of e: $\lim_{x\to\infty} \left(1+\frac{1}{x}\right)^x$

In this case, r is the interest rate (as a decimal) and n is the number of times or 'periods' in a year the invested amount will be compounded (for example, if the invested amount were compounded each day of the year, n would equal 365) Now lets substitute $x=\frac{n}{r}$. The formula now becomes:

$\left(1+\frac{1}{x}\right)^{xr}$

Based on the similarities between this formula and the evaluation of e, the formula can be rewritten as $e^r$, which is the first term in the most recognised variation of the formula for continuous interest, $e^r - 1$, but the question I have is: since we incorporate e into the formula for continuous interest, doesn't that mean we assume that the interest rate is always 100%? Since 100% = 1 (as a decimal), and using the evaluation of e, r is always equal to 1. Hence, why is it that this assumption is correct? Or is my logic flawed?

• I do not see what forces r=1. You can use the identity $(1+1/x)^{xr}=((1+1/x)^x)^r$ and then take the limit. Also, these formulas compute the factor by which you multiply the principal to get the total after interest, not the interest itself. – Ian Jul 18 '17 at 12:38
• youtube.com/watch?v=AuA2EAgAegE may help – user451844 Jul 18 '17 at 12:45

First, $(1+{1\over x})^x$ is not the same thing as $\lim_{x\rightarrow\infty}(1+{1\over x})^x$, so $e$ is not the base of your formula.

Second, continuous interest is not $(1+{r\over n})^n$; this is actually an improperly written formula for discrete-time compound interest factor. It should be correctly expressed as $(1+{r\over n})^{nt}$, where discrete time is $t$ and $n$ is compounding frequency. This expression can nevertheless be evaluated even if $t$ is continuous.

However, they're still related. The formula for proper continuous compound interest factor is $e^{rt}$. This is the consequence of

$$\lim_{n\rightarrow\infty}(1+{r\over n})^{nt}=e^{rt}$$

In words, as the compounding frequency becomes infinitely large such that the interest is compounded every infinitesimally small amount of time, then the Euler's constant actually comes into play. This feature is not by construction, but rather just an artifact of how we calculate interest, but it does make generalizing compound interest in continuous time easy.

To check if it's true, we can look at the instantaneous rate of return of $e^{rt}$:

$${d\over dt}\ln(e^{rt})={d\over dt}rt=r$$

It flips back to the interest rate, which is the rate of return.

• Your answer makes complete sense apart from 2 areas: the first of which being why $nt$ is used as opposed to simply $t$, I'm assuming it's to calculate the total sum of compounding periods, but why is this necessary? The second area being, can you elaborate on how $e^{rt}$ is the consequence of $\lim_{n\to\infty}\left(1+\frac{r}{n}\right)^{nt}$? Thank you. – joshuaheckroodt Jul 19 '17 at 1:10
• For your first question, this because in real life situations, interest is compounded on a more frequent base than the nominal interest rate. For example, if your annual interest rate is 6%, and interest rate is compounded monthly, so your nominal interest rate is divided by 12 and compounded for 12 times in a year, therefore you must raise the base $(1+r/n)$, which is the principle plus interest at the monthly rate to the power of $nt$, even if $t$ is just 1 year, you've compounded by $n=12$ times. – seamux Jul 19 '17 at 8:27
• For your second question, you already know that $e=\lim_{x\rightarrow\infty}(1+1/x)^x=e$, an extension of that is $e=\lim_{x\rightarrow\infty}(1+a/x)^x=e^a$. The proof of this is beyond the scope of this answer but you can look it up. Now because $e^{rt}=(e^r)^t$, you get $\lim_{n\rightarrow\infty}(1+r/n)^{nt}=e^rt$. – seamux Jul 19 '17 at 8:33
• By the way, the answer to your first question is also why there is a difference between annual interest rate and annual percentage rate of change (APR) that you often see on bank statements. – seamux Jul 19 '17 at 8:36
• Just noticed the typos... ignore the two $e=$'s before the limits. – seamux Jul 19 '17 at 15:58

Yes, the numerator in the limit for $e$ is $1$ rather than $r$ -- but note that the denominator is no longer $n$, so you have no particular reason to expect that the numerator in that fraction can be interpreted as (or ought to coincide with) a yearly interest rate.