I was wondering

mattflint50

Junior Member
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Apr 25, 2005
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I am curious to know why the lim as n approaches infinity of P(1+I/N)^N turns into Y=P(e^IT) when it is used continuously. This would clear things up for me. THanks
 
Is it P times, \(\displaystyle \L
P \times \left( {1 + \frac{I}{N}} \right)^N\)? If so, the answer is yes.

Or is P a function: \(\displaystyle \L
P\left( {1 + \frac{I}{N}} \right)^N = P^N \left( {1 + \frac{I}{N}} \right)\) In the case we need to know more about P.
 
Hello, mattflint501

I am curious to know why limnP(1+in)n\displaystyle \:\lim_{n\to\infty}P\left(1\,+\,\frac{i}{n}\right)^n turns into \(\displaystyle y\:=\:p\cdot e^{iT}\) 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: \(\displaystyle \L\,\lim_{z\to\infty}\left(1\,+\,\frac{1}{z}\right)^z\;=\;e\)

    \displaystyle \;\;Note that the three z\displaystyle z's must be identical.


Given a compound interest problem with principal P, annual interest rate i\displaystyle \text{Given a compound interest problem with principal }P\text{, annual interest rate }i
    for T years with n compounding periods per year,\displaystyle \;\;\text{for }T\text{ years with }n\text{ 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 ni\displaystyle \text{We note that, if }n\to\infty\text{, then }\frac{n}{i}\to \infty


\(\displaystyle \text{Let }z\,=\,\frac{n}{i}\text{ and we have: }\L\,y\;=\;P\left[\left(1\,+\,\frac{1}{z}\right)^z\right]^{iT}\)

\(\displaystyle \text{Let }z\to\infty:\L\;y\;=\;\lim_{z\to\infty}\left[P\left(1\,+\,\frac{1}{z}\right)^z\right]^{iT}\)

\(\displaystyle \text{and we have: }\L\:y\;=\;P\left[\underbrace{\lim_{z\to\infty}\left(1\,+\,\frac{1}{z}\right)^z}\right]^{iT}\)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . This is e\displaystyle e !

Therefore: \(\displaystyle \L\:y\;=\;P\cdot e^{iT}\)
 
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