- #1
QuarkCharmer
- 1,051
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I should have made the title say "using L'h..." rather than what it is, I apologize.
Prove the compounding interest formula from the following:
[tex]A=A_{0}(a+\frac{r}{n})^{nt}[/tex]
[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.
Homework Statement
Prove the compounding interest formula from the following:
[tex]A=A_{0}(a+\frac{r}{n})^{nt}[/tex]
Homework Equations
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.
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