1. ## Proof of e

Hi MHF.

I would like to see some proof/detailed explanation for $H\cdot \left(\frac{i}{m}\right)^{mn}=He^{in}$

Where H is the starting amount, i is the percentage and m is the # of times per year where growth is calculated and n is the number of years. e is Euler's number.

2. ## Re: Proof of e

The formula you want, in terms of the variables you have defined, is:

$\lim_{m\to\infty}\left(H\left(1+\frac{i}{m} \right)^{mn} \right)=He^{in}$

To prove this statement requires some methods usually found in a second semester of elementary calculus. To begin, let's write:

$\lim_{m\to\infty}\left(H\left(1+ \frac{i}{m} \right)^{mn} \right)=L$

Because the constant $H$ does not depend on $m$, we may divide through by this to obtain:

$\lim_{m\to\infty}\left(\left(1+ \frac{i}{m} \right)^{mn} \right)=\frac{L}{H}$

Now, taking the natural log of both sides, we have:

$\ln\left(\lim_{m\to\infty}\left(\left(1+ \frac{i}{m} \right)^{mn} \right) \right)=\ln\left(\frac{L}{H} \right)$

The log of a limit is equal to the limit of a log, hence:

$\lim_{m\to\infty}\left(\ln\left(\left(1+ \frac{i}{m} \right)^{mn} \right) \right)=\ln\left(\frac{L}{H} \right)$

Applying the property of logs $\log_a\left(b^c \right)=c\cdot\log_a(b)$ we have:

$\lim_{m\to\infty}\left(mn\cdot\ln\left(\left(1+ \frac{i}{m} \right) \right) \right)=\ln\left(\frac{L}{H} \right)$

Dividing through by $n$:

$\lim_{m\to\infty}\left(m\cdot\ln\left(\left(1+ \frac{i}{m} \right) \right) \right)=\frac{\ln\left(\frac{L}{H} \right)}{n}$

Bring $m$ to the denominator:

$\lim_{m\to\infty}\left(\frac{\ln\left(\left(1+ \frac{i}{m} \right)}{\frac{1}{m}} \right) \right)=\frac{\ln\left(\frac{L}{H} \right)}{n}$

We now have the indeterminate form 0/0, thus application of L'Hôpital's rule reveals:

$\lim_{m\to\infty}\left(\frac{-\frac{i}{\left(1+\frac{i}{m} \right)m^2}}{-\frac{1}{m^2}} \right) \right)=\frac{\ln\left(\frac{L}{H} \right)}{n}$

Simplify:

$\lim_{m\to\infty}\left(\frac{i}{1+\frac{i}{m}} \right)=\frac{\ln\left(\frac{L}{H} \right)}{n}$

We see now the term involving $m$ vanishes to zero, hence:

$i=\frac{\ln\left(\frac{L}{H} \right)}{n}$

Solve for $L$

$in=\ln\left(\frac{L}{H} \right)$

$\frac{L}{H}=e^{in}$

$L=He^{in}$

3. ## Re: Proof of e

Thank you very much.