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:

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