# Compound Interest

(and L'Hopital's rule -- Sometimes limits are kind of tricky....)

## Derivation of the Compound Interest Formula

In financial mathematics we tend to use an unusual base for an exponential:

$P(t)=P_{0}\left(1+{\frac {r}{n}}\right)^{nt}$ Let's see where this model comes from: it's the compound interest model -- that is, interest on top of interest. If you receive simple interest at a rate $r$ on an initial investment $P_{0}$ , then at the end of the year you'd have

$P(1)=P_{0}+r\cdot {P_{0}}=P_{0}(1+r)$ If the interest is truly simple, then you simply accumulate $rP_{0}$ at the end of each year, for a total (after $t$ years) of $P(t)=P_{0}\left(1+tr\right)$ .

If, on the other hand, you keep re-investing the money accrued, then you have what is known as "annual compounding": the formular for it would be

$P(t)=P_{0}\left(1+r\right)^{t}$ Now, in compound interest, you take your interest rate, split it up several (say 12) ways, and then do the computation several (e.g. 12) times:

$P(1/12)=P_{0}+(r/12)\cdot {P_{0}}=P_{0}(1+r/12)$ So that

$P(1)=P_{0}\left(1+r/12\right)^{12}$ And if you did it for $t$ years, you'd have

$P(t)=P_{0}\left(1+r/12\right)^{12t}$ Hence, more generally,

$P(t)=P_{0}\left(1+{\frac {r}{n}}\right)^{nt}$ You may have encountered a few laws such as The rule of 72 (70, 71, 69.3,...) to calculate doubling time: forget them! Just do the math. The question is this: for what value of t is

$P(t)=P_{0}\left(1+{\frac {r}{n}}\right)^{nt}=2P_{0}$ ?

Solving

$P_{0}\left(1+{\frac {r}{n}}\right)^{nt}=2P_{0}$ for t, we get

$t={\frac {\ln 2}{n\ln(1+{\frac {r}{n}})}}$ where n is the number of compoundings per year, and r is given as a decimal (e.g. 9% is represented by .09). This is the doubling time.

When compounding is continuous (i.e. $n\to \infty$ ), this reduces to the very lovely rule

${\frac {\ln 2}{r}}$ Now, how do we know that

$\lim _{n\to \infty }{\frac {\ln 2}{n\ln(1+{\frac {r}{n}})}}={\frac {\ln 2}{r}}$ ?

That is, how do we know that

$\lim _{n\to \infty }\ {n\ln(1+{\frac {r}{n}})}={r}$ ?

The answer, of course, is L'Hopital's Rule, which is useful in solving certain indeterminate limits. Let's rewrite it a little: we want to show that

$\lim _{n\to \infty }{\frac {\ln(1+{\frac {r}{n}})}{\frac {1}{n}}}={r}$ Here's the general situation: http://www.nku.edu/~longa/classes/mat122/days/day35/thm7-7-1.gif

And if the limit of the quotient of the derivatives is indeterminate, then iterate -- that is, do it again! Then work with the ratio of the second (or higher) derivatives.

## Motivation of L'Hopital's Rule

(requires the limit definition of the derivative).

Let's suppose that we're looking at an indeterminate limit

$\lim _{x\to {a}}{\frac {f(x)}{g(x)}}$ of the form 0/0. We're assuming that both functions are differentiable at $\left.a\right.$ , which means that both functions are continuous at $\left.a\right.$ (i.e. $\left.f(a)=g(a)=0\right.$ ). Let's also assume that $\left.g'(a)\neq {0}\right.$ .

We start with the limit of the derivatives, and make our way over to the original limit:

$\lim _{x\to {a}}{\frac {f'(x)}{g'(x)}}={\frac {f'(a)}{g'(a)}}={\frac {\lim _{x\to {a}}{\frac {f(x)-f(a)}{x-a}}}{\lim _{x\to {a}}{\frac {g(x)-g(a)}{x-a}}}}=\lim _{x\to {a}}{\frac {\frac {f(x)-f(a)}{x-a}}{\frac {g(x)-g(a)}{x-a}}}=\lim _{x\to {a}}{\frac {f(x)-f(a)}{g(x)-g(a)}}=\lim _{x\to {a}}{\frac {f(x)}{g(x)}}$ 