5.4 ** The continuous compounding formula derivation

Where does the continuous compounding formula come from?

$$\fbox{$\lim_{m \rightarrow \infty} P_0 \left(1 + \frac{r}{m}\right)^{m\, t} = P_0 \, e^{r\,t}.$ }$$

Assume the limit exists, and call it L, then:

$$L = \lim_{m \rightarrow \infty} P_0 \left(1 + \frac{r}{m}\right)^{m\, t}.$$

So

$$ln(L) = ln \left(\lim_{m \rightarrow \infty} P_0 \left(1 + \frac{r}{m}\right)^{m\, t}\right).$$

If we are allowed ...

$$ln(L) = \lim_{m \rightarrow \infty} ln\left(P_0 \left(1 + \frac{r}{m}\right)^{m\, t}\right).$$

Now, log of a product is the sum of the logs ...

$$ln(L) = \lim_{m \rightarrow \infty}\left(ln(P_0) + ln \left[\left(1 + \frac{r}{m}\right)^{m\, t}\right]\right).$$

Use log rules:

$$ln(L) = \lim_{m \rightarrow \infty}\left(ln(P_0) +(m\,t)\, ln \left(1 + \frac{r}{m}\right)\right).$$

But as m gets large, so $\frac{r}{m}$ gets really small, so can use the log approximation $ln(1 + h) \sim h$, to get

$$ln(L) = \lim_{m \rightarrow \infty}\left(ln(P_0) +(m\,t)\, \frac{r}{m}\right).$$

Cancel to get

$$ln(L) = \lim_{m\rightarrow \infty}\left(ln(P_0) + r\,t \right).$$

Now exp both sides to get

$$L = P_0\, e^{r\,t}.$$