0

I understand that one way of understanding the meaning of the number $e$ is to form a compound interest formula, $A = \left(1+\frac{1}{n}\right)^{nx}$ and then let $n\rightarrow \infty$ for which this converges to $e$. However, this occurs by setting the principle to 1, the rate to 1, and the time to $x$. And I see how you can manipulate this to get any other principle--you just multiply by the principle you want. So $Pe^{x}$ would be any other principle. But how do you get any other rate?

That is to say, suppose you have $A = \left(1+\frac{r}{n}\right)^{nx}$ and let $n\rightarrow \infty$. To what does this converge?

Addem
  • 5,656

2 Answers2

1

Recall that one of the definitions of the exponential function is: $$e^{r}:=\lim\limits_{n\to\infty}\left(1+\dfrac rn\right)^{n}.$$ If we raise both sides to $x$ we get: $$e^{rx}=\lim\limits_{n\to\infty}\left(1+\dfrac rn\right)^{xn}.$$

Hakim
  • 10,213
1

It converges to $$e^{rx}$$

To see this you can write $$A = \left(1+\frac{r}{n}\right)^{nx} = \left(1+\frac{1}{\frac{n}{r}}\right)^{\frac{n}{r}xr} \to e^{xr}$$

Ant
  • 21,098