- #1
McFluffy
- 37
- 1
Homework Statement
The constant ##e## can be defined in many ways. My first exposure to this number involved compound interests. Specifically, if you decide to continuously compound 1$ at a 100% interest rate for a certain period of time(year, month, etc), you'll end up with ##e##$ where it is defined as: ##\lim_{n\to\infty}(1+\frac{100\%}{n})^{n}=e##.
I've been told that if you wanted to continuously compound some amount of money, ##P## at some interest rate per unit time, ##r## and that you want to do it ##t## units of time you'll get: ##P\cdot e^{rt}##. I understand why there's an exponent ##t## there because of how multiplication is defined but what I don't understand is why is ##r## there. I'm trying to do this using only basic properties about limits and not other tools from calculus if possible as I'm not well adept at it yet. So the problem statement that I asked myself is:
How do you show that ##e^{r} = \lim_{n\to\infty} (1+\frac{r}{n})^n##?
Homework Equations
##\lim_{n\to\infty}(1+\frac{1}{n})^n =e##
##\sum_{n=0}^{\infty} \frac{1}{n!} = e##
## (a+b)^{n} = \sum_{k=0}^{n}## ## n \choose {k}## ##a^{n-k}b^{k}##,
3. The Attempt at a Solution
I tried expanding ##(1+\frac{r}{n})^n## using the binomial theorem to show that it is ##e^{r}## from there using its other definition which ##\sum_{n=0}^{\infty} \frac{1}{n!} ## but I all I got was useless manipulations of the equation with tedious writings and was pretty much lost on how to get to there.