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Thread: Proof of e

  1. #1
    Senior Member Paze's Avatar
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    Proof of e

    Hi MHF.

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

    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.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Proof of e

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

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

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

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

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

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

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

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

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

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

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

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

    Dividing through by $\displaystyle n$:

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

    Bring $\displaystyle m$ to the denominator:

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

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

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

    Simplify:

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

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

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

    Solve for $\displaystyle L$

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

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

    $\displaystyle L=He^{in}$
    Thanks from Paze and agentmulder
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  3. #3
    Senior Member Paze's Avatar
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    Re: Proof of e

    Thank you very much.
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