Difference between revisions of "Talk:Interesting integration MA181Fall2008bell" - Rhea

Revision as of 10:00, 3 October 2008

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