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Please check my proof (L'hôpital's rule)

  1. Sep 15, 2011 #1
    I should have made the title say "using L'h...." rather than what it is, I apologize.

    1. The problem statement, all variables and given/known data
    Prove the compounding interest formula from the following:
    [tex]A=A_{0}(a+\frac{r}{n})^{nt}[/tex]

    2. Relevant equations

    3. The attempt at a solution

    [tex]A=A_{0}(a+\frac{r}{n})^{nt}[/tex]

    [tex]lim_{n\to\infty}A_{0}(a+\frac{r}{n})^{nt}[/tex]

    [tex]A_{0}e^{t lim_{n\to\infty}n ln(a+\frac{r}{n})}[/tex]

    let u = [itex]lim_{n\to\infty}n ln(a+\frac{r}{n})[/itex]

    So now the desired function is [itex]A = A_{0}e^{tu}[/itex]

    [tex]u = lim_{n\to\infty} \frac{ln(1+\frac{r}{n})}{\frac{1}{n}}[/tex]

    Using L'hôpital's rule gives that:

    [tex]u = lim_{n\to\infty} \frac{\frac{d}{dn}ln(1+\frac{r}{n})}{\frac{d}{dn} \frac{1}{n}}[/tex]

    [tex]u = lim_{n\to\infty}\frac{rn^{2}}{n^{2}+rn}[/tex]

    [tex]u = lim_{n\to\infty}\frac{\frac{rn^{2}}{n^{2}}}{\frac{n^{2}}{n^{2}}+\frac{rn}{n^{2}}}[/tex]

    [tex]u = lim_{n\to\infty} \frac{r}{1+\frac{r}{n}} = \frac{r}{1} = r[/tex]

    [tex]u = r[/tex]

    Now, re-substituting r back in for u yields:

    [tex]A = A_{0}e^{rt}[/tex]


    Which is the typical [itex]Pe^{rt}[/itex] formula we all know and love?

    I don't want to look it up and spoil it because I think I am right. Not sure if you can raise e to the ln of something, but I don't see why not. It's not like it changes anything and I figure you have to get an e in there somehow.
     
    Last edited: Sep 15, 2011
  2. jcsd
  3. Sep 15, 2011 #2

    micromass

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    Some remarks:

    Why did you drop the a?? You have exchanged a by 1. Eventually it won't matter, but I don't see how you can drop it just like that.

    This is correct, but I don't see where you get it from.

    Yes, you can do it, and it's one of the great tricks in working with limits. Did you discover that trick all by yourself?? Great!!
     
  4. Sep 15, 2011 #3

    vela

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    Why did you set a=1?
     
  5. Sep 15, 2011 #4
    Oh wow, yeah that "a" should have been a "1" from the start. I typed out the formula I was starting with incorrectly and then copy/paste the errors further into the problem until I finally typed a line by hand and fixed it!

    I'm a horrible latex-er?

    Truth be told, I didn't want to type out the intermediate steps where I took the derivative and simplified. I know that step to be correct and this is not a real "proof" that I am turning in to anyone. I am just doing some extra problems from Stuart 6E 7.8 (This one is #85).

    Aside from the missing intermediate steps, and the horrid typo (I swear it's correct on my handwritten paper), everything else is okay? More specifically, if I were to document all the steps in a proof based writing (which I have no idea how to do), it would make sense how I came to the conclusion?

    Thanks for all the input.
     
  6. Sep 15, 2011 #5

    micromass

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    Yes, everything is clear and correct!
     
  7. Sep 15, 2011 #6
    I thought so but I doubt it. I think I remember a similar tactic used in a proof of something on khanacademy and I just stole the idea.

    Thanks for the help!
     
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