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'''p*Ei(x(log(r)-log(p)))''' | '''p*Ei(x(log(r)-log(p)))''' | ||
| − | Note: <math> Ei(x) = \int_{ | + | Note: <math> Ei(x) = \int_{x}^{\infty} \frac{e^{-t}}{t} dt </math> |
I don't know how to use this integral, but I did some manipulation and got this: | I don't know how to use this integral, but I did some manipulation and got this: | ||
<math> Total = \frac{A[(r+1)^{t+1}-(r+1)]}{r} </math> | <math> Total = \frac{A[(r+1)^{t+1}-(r+1)]}{r} </math> | ||
| + | |||
| + | or... | ||
| + | |||
| + | <math> Total = \frac{A}{i}e^{(t+1)\ln{(i+1)}}-\frac{A(i+1)}{i} </math> | ||
| + | |||
| + | See discussion for more info... | ||
| + | |||
| + | The above integral is incorrect. i equals r/p. I forgot to add the investment again and again at each period. Otherwise, I should have got a simple exponential I guess. --[[User:Gbrizend|Gary Brizendine II]] | ||
| + | ---[[User:Gbrizend|Gary Brizendine II]] | ||
Latest revision as of 09:43, 7 October 2008
First off, this is not part of homework. This equation (if I did it right) is the summation of an investment 'A' that gains interest over period 'p' and time 't' in years. The investment 'A' is added every period. I originally got $ A \sum_{t=0}^{n} \frac{r^t}{p^{t}} dt $, but I wanted to turn it into an integral and pulled out a $ \frac{t}{p} $ so I would have a dt. That led me to the integral below. Does it make sense and does anyone know how to integrate the problem?
Integrate this:
$ A \int_{0}^{n} \frac{r^t}{t*p^{t-1}} dt $
I searched how to do it on matlab, but could not find it. Then, I found this website on Wolfram. It integrates it using mathematica. Here is what it got:
p*Ei(x(log(r)-log(p)))
Note: $ Ei(x) = \int_{x}^{\infty} \frac{e^{-t}}{t} dt $
I don't know how to use this integral, but I did some manipulation and got this:
$ Total = \frac{A[(r+1)^{t+1}-(r+1)]}{r} $
or...
$ Total = \frac{A}{i}e^{(t+1)\ln{(i+1)}}-\frac{A(i+1)}{i} $
See discussion for more info...
The above integral is incorrect. i equals r/p. I forgot to add the investment again and again at each period. Otherwise, I should have got a simple exponential I guess. --Gary Brizendine II ---Gary Brizendine II
