I'm not sure that your original function makes that much sense; I can't say that I can tell how you got to that point.
Just checking out how normal compounded interest works, I checked Wikipedia and rediscovered the formula:
$ A=P\bigg(1+\frac{r}{n}\bigg)^{nt} $
P = principal amount (initial investment)
r = annual nominal interest rate (as a decimal)
n = number of times the interest is compounded per year
t = number of years
A = amount after time t
I've tried to break that down into good 'ole
$ A=Pe^{rt} $
but haven't had any luck; I always end up with an indeterminate value relating one and infinity that I can't break down into something manageable with L'Hopital's.--Jmason 15:53, 2 October 2008 (UTC)
- To Gary: Yeah, I have to agree with John, I'm not following your math here. Could you tell us how you got to that equation?
To John: The derivation is actually in our book, but I'll redo the work here to practice with Latex and so you can see it here instead of going to the book.
$ A=P\bigg(1+\frac{r}{n}\bigg)^{nt} $
Now we we want to find the limit as n goes to infinity, since we're trying to compound the interest continuously, and therefore have an infinite number of times we compound.
$ P\lim_{n\to\infty}\bigg(1+\frac{r}{n}\bigg)^{nt} $
We now have a limit in the indeterminate form $ 1^{\infty} $
Now I'm going to stray from what the book showed us and use L'H rule to show that the limit approaches $ Pe^{rt} $
First, drop the P, we can multiply it back in later. Also, since it's in form $ 1^{\infty} $ we can try to find its limit by taking the natural log of it and its limit (Which means when we find the new limit, we have to raise e to that power to get the right limit, as you already know.)
$ \lim_{n\to\infty}\ln{\bigg(1+\frac{r}{n}\bigg)^{nt}} $
Move the nt to the front
$ \lim_{n\to\infty}nt\ln{\bigg(1+\frac{r}{n}\bigg)} $
And move the nt to the bottom by inverting it
$ \lim_{n\to\infty}\frac{\ln{\bigg(1+\frac{r}{n}\bigg)}}{\frac{1}{nt}} $
Which is now in the indeterminate form $ \frac{0}{0} $ So apply L'H rule and find derivatives of top and bottom functions:
$ \lim_{n\to\infty}\frac{\frac{-r}{(1+\frac{r}{n})n^2}}{\frac{-1}{n^2t}} $
Now the $ -n^2 $ cancel and we can take the limit as n approaches infinity.
$ \lim_{n\to\infty}\frac{rt}{1+\frac{r}{n}}=\frac{rt}{1}=rt $
Now take e to this power to get the actual limit.
$ \lim_{n\to\infty}\bigg(1+\frac{r}{n}\bigg)^{nt}=e^{rt} $
And now we simply add back in our constant to solve completely.
$ A=P\lim_{n\to\infty}\bigg(1+\frac{r}{n}\bigg)^{nt}=Pe^{rt} $ Jhunsber
What I am doing is calculating a sum where you initially invest an amount, and then you invest that same amount again next period. For instance, I invest $1000 in an IRA for year one. Year 2, I invest another $1000 on top that. Year 3, I invest another $1000.
It looks like this:
A= Amount Invested each period
p= period
r= annual percentage rate (in this case it is positive for an investment)
t= time in years
Period(1) = $ A(\frac{r}{p} +1) = \frac{Ar}{p} +A $
Period(2) = $ [Year(1)+A](\frac{r}{p} +1) = \frac{Ar^2}{p^2}+\frac{3Ar}{p} +2A $
Period(3) = ...... ---Gary Brizendine II
Notice...
i= r/p = periodic interest rate
P(1) = A(i+1)
P(2) = A(i+1)(i+2)
P(3) = A(i+1)(i^2+3i+3)
P(4) = A(i+1)(i^3+4i^2+4i+4)
so.....
$ P(t) = -(t-1)(\frac{r}{p})^{t-1} + At(\frac{r}{p}+1) \sum_{t=1}{N}(\frac{r}{p})^{t-1} $
I'll leave what I just put up there, but it is wrong. I noticed the coefficients inside the powers of 'i' parentheses. I assumed for the fourth period that they would be fours. I was wrong. I carried it through to five and noticed something interesting. It creates a pascal's triangle effect. The fourth is 1-4-6-4; the fifth is 1-5-10-10-5; the sixth is 1-6-15-20-15-6. This makes things a little more complicated. The equation I came up with on the main page without sums or integrals is correct. I got this by solving for a geometric series. I will try to solve for this new pascal configuration. --Gary Brizendine II
Wow!!!! Solving it like this actually gave me the exact formula I got using series. I guess that you can not use a summation here. For some reason, I thought you could because I got a sigma in there. What did I learn from this?: just because you have a sigma doesn't mean you can necessarily have an integral. Correct me if I'm wrong.
Formula:
$ Total = \frac{A[(i+1)^{t+1}-(i+1)]}{i} $
This can also be:
$ Total = \frac{A[(e)^{(t+1)\ln{(i+1)}}-(i+1)]}{i} $
This makes the exponential function:
$ Total = Ae^{(t+1)\ln{(i+1)}}-\frac{A(i+1)}{i} $
Thanks for the comments! --Gary Brizendine II
