Hello, mattflint501
I am curious to know why
n→∞limP(1+ni)n turns into \(\displaystyle y\:=\

\cdot e^{iT}\) when it is used continuously.
This would clear things up for me. Thanks
Do you
really want to know?
It's a long and nasty derivation, usually found in your textbook.
It is based on the defintion: \(\displaystyle \L\,\lim_{z\to\infty}\left(1\,+\,\frac{1}{z}\right)^z\;=\;e\)
Note that the three
z's must be identical.
Given a compound interest problem with principal P, annual interest rate i
for T years with n compounding periods per year,
\(\displaystyle \;\;\;\text{the final amount }y\text{ is: }\L\:y\;=\;P\left(1\,+\,\frac{i}{n}\right)^{Tn}\)
\(\displaystyle \text{"Invert" that fraction: }\L\:y\;=\;P\left(1\,+\,\frac{1}{\left(\frac{n}{i}\right)}\right)^{Tn}\)
\(\displaystyle \text{Multiply the exponent by }\frac{i}{i}\,:\L\;\;y\;=\;\left(1\,+\,\frac{1}{\left(\frac{n}{i}\right)}\right)^{Tn\,\cdot\,\frac{i}{i}} \;=\;P\left(1\,+\,\frac{1}{\left(\frac{n}{1}\right)}\right)^{Ti\left(\frac{n}{i}\right)}\)
\(\displaystyle \text{And we have: }\L\:y\;=\;P\left[\left(1\,+\,\frac{1}{\left(\frac{n}{i}\right)}\right)^{\frac{n}{i}}\right]^{iT}\)
We note that, if n→∞, then in→∞
\(\displaystyle \text{Let }z\,=\,\frac{n}{i}\text{ and we have: }\L\,y\;=\;P\left[\left(1\,+\,\frac{1}{z}\right)^z\right]^{iT}\)
\(\displaystyle \text{Let }z\to\infty:\L\;y\;=\;\lim_{z\to\infty}\left[P\left(1\,+\,\frac{1}{z}\right)^z\right]^{iT}\)
\(\displaystyle \text{and we have: }\L\:y\;=\;P\left[\underbrace{\lim_{z\to\infty}\left(1\,+\,\frac{1}{z}\right)^z}\right]^{iT}\)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . This is
e !
Therefore: \(\displaystyle \L\:y\;=\;P\cdot e^{iT}\)