mattflint50
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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
I was wondering
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pka
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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.
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
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 \(\displaystyle z\)'s must be identical.
\(\displaystyle \text{Given a compound interest problem with principal }P\text{, annual interest rate }i\)
\(\displaystyle \;\;\text{for }T\text{ years with }n\text{ compounding periods per year,}\)
\(\displaystyle \;\;\;\text{the final amount }y\text{ is: }\L\:y\;=\;P\left(1\,+\,\frac{i}{n}\right)^{Tn}\)
\(\displaystyle \text{"Invert" that fraction: }\L\:y\;=\;P\left(1\,+\,\frac{1}{\left(\frac{n}{i}\right)}\right)^{Tn}\)
\(\displaystyle \text{Multiply the exponent by }\frac{i}{i}\,:\L\;\;y\;=\;\left(1\,+\,\frac{1}{\left(\frac{n}{i}\right)}\right)^{Tn\,\cdot\,\frac{i}{i}} \;=\;P\left(1\,+\,\frac{1}{\left(\frac{n}{1}\right)}\right)^{Ti\left(\frac{n}{i}\right)}\)
\(\displaystyle \text{And we have: }\L\:y\;=\;P\left[\left(1\,+\,\frac{1}{\left(\frac{n}{i}\right)}\right)^{\frac{n}{i}}\right]^{iT}\)
\(\displaystyle \text{We note that, if }n\to\infty\text{, then }\frac{n}{i}\to \infty\)
\(\displaystyle \text{Let }z\,=\,\frac{n}{i}\text{ and we have: }\L\,y\;=\;P\left[\left(1\,+\,\frac{1}{z}\right)^z\right]^{iT}\)
\(\displaystyle \text{Let }z\to\infty:\L\;y\;=\;\lim_{z\to\infty}\left[P\left(1\,+\,\frac{1}{z}\right)^z\right]^{iT}\)
\(\displaystyle \text{and we have: }\L\:y\;=\;P\left[\underbrace{\lim_{z\to\infty}\left(1\,+\,\frac{1}{z}\right)^z}\right]^{iT}\)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . This is \(\displaystyle e\) !
Therefore: \(\displaystyle \L\:y\;=\;P\cdot e^{iT}\)
Do you really want to know?I am curious to know why \(\displaystyle \:\lim_{n\to\infty}P\left(1\,+\,\frac{i}{n}\right)^n\) turns into \(\displaystyle y\:=\\cdot e^{iT}\) when it is used continuously.
This would clear things up for me. Thanks
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 \(\displaystyle z\)'s must be identical.
\(\displaystyle \text{Given a compound interest problem with principal }P\text{, annual interest rate }i\)
\(\displaystyle \;\;\text{for }T\text{ years with }n\text{ compounding periods per year,}\)
\(\displaystyle \;\;\;\text{the final amount }y\text{ is: }\L\:y\;=\;P\left(1\,+\,\frac{i}{n}\right)^{Tn}\)
\(\displaystyle \text{"Invert" that fraction: }\L\:y\;=\;P\left(1\,+\,\frac{1}{\left(\frac{n}{i}\right)}\right)^{Tn}\)
\(\displaystyle \text{Multiply the exponent by }\frac{i}{i}\,:\L\;\;y\;=\;\left(1\,+\,\frac{1}{\left(\frac{n}{i}\right)}\right)^{Tn\,\cdot\,\frac{i}{i}} \;=\;P\left(1\,+\,\frac{1}{\left(\frac{n}{1}\right)}\right)^{Ti\left(\frac{n}{i}\right)}\)
\(\displaystyle \text{And we have: }\L\:y\;=\;P\left[\left(1\,+\,\frac{1}{\left(\frac{n}{i}\right)}\right)^{\frac{n}{i}}\right]^{iT}\)
\(\displaystyle \text{We note that, if }n\to\infty\text{, then }\frac{n}{i}\to \infty\)
\(\displaystyle \text{Let }z\,=\,\frac{n}{i}\text{ and we have: }\L\,y\;=\;P\left[\left(1\,+\,\frac{1}{z}\right)^z\right]^{iT}\)
\(\displaystyle \text{Let }z\to\infty:\L\;y\;=\;\lim_{z\to\infty}\left[P\left(1\,+\,\frac{1}{z}\right)^z\right]^{iT}\)
\(\displaystyle \text{and we have: }\L\:y\;=\;P\left[\underbrace{\lim_{z\to\infty}\left(1\,+\,\frac{1}{z}\right)^z}\right]^{iT}\)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . This is \(\displaystyle e\) !
Therefore: \(\displaystyle \L\:y\;=\;P\cdot e^{iT}\)