1. ## Continuous compounding formula

$\displaystyle F = Ne^{kt}$

I was reading up on this formula, and alot of the examples given are usually bank interest problems. I've noticed that they will always mention that the rates are continuously compounding, but I haven't really found a definition for that. One thing I read said continuous compounding was just taking the interest as part of the principle (which is the general definition of compounding interest). Another definition I read said that the compounding occurs "immeasurably fast." I've noticed the formula itself doesn't actually let you specify the when compounding occurs, but rather only the amount of time that has passed.

Here's an example problem I read:

You have 1000 grams of a radioactive isotope. The half-life of this isotope is 8 days. How many grams will remain after 30 days?

The first thing that comes to mind is that the rate is 0.5 for a duration of 8 days, but it states that this is incorrect and that the rate has to be calculated with the rest of the information given. I did that and came up with a fairly long negative decimal number. The number cleraly works with the formula, but is that rate only useful for that formula? I guess this means that for this formula you would have to calculate the rate every time unless the problem specifically states that the rate given to you is continously compounding?

Here's a problem I made up:

You deposit $500 into an account. You recieve a compound interest rate of 5% every 3 months. How much will you have after 1 year? Doing the problem manually, I got the answer of 607.753125. I tried fooling around with the formula to see if I could replicate that:$\displaystyle F = 500e^{(0.05)(3)}$Nope, didn't work.$\displaystyle F = 500e^{(0.05)(12)}$This doesn't even feel like it makes sense.$\displaystyle F = 500e^{(0.05)(4)}$This yields 610.701379, which is pretty close. Is it just a coincidence? The only relation I see is that 4 is the number of times the compounding happens in 1 year, but the formula asks for the total duration. I went back to the radioactive isotope problem: If the half-life is 8 days, and the duration is 30 days, the number of times compounding will happen is 30/8 or 3.75. I have a 1000 grams and I need to know how much is left after 30 days.$\displaystyle F = 500e^{(-0.5)(3.75)}$This yields about 77, which is close to the other answer of 74 I got when I calculated the rate using the given information. So... am I right that this formula requires you to calculate the rate every time (unless given a rate you know is appropriate for this formula)? Is it wrong to use the formula in the way I tested above? It gets close answers, but they aren't the same. 2. Originally Posted by Phire You have 1000 grams of a radioactive isotope. The half-life of this isotope is 8 days. How many grams will remain after 30 days? This one is exponential decay and is a different type of equation to compound interested. For this one you'd use the two main formulae in exponential decay:$\displaystyle A(t) = A_0e^{-\lambda t}$(eq1)$\displaystyle \lambda = \frac{ln(2)}{t_{1/2}}$(eq2) These can be combined into$\displaystyle A(t) = A_0e^{-\frac{tln(2)}{t_{1/2}}}$Where •$\displaystyle A(t)$is amount left at time t •$\displaystyle A_0$is initial amount at t=0 •$\displaystyle \lambda$is the decay constant •$\displaystyle t$is time Here's a problem I made up: You deposit$500 into an account. You recieve a compound interest rate of 5% every 3 months. How much will you have after 1 year?
For compound interest you can't really use e as a base because it is not strictly accurate:

$\displaystyle \lim_{n\to\infty} \left( 1 + \frac{1}{n} \right)^n = e$

so using e is only accurate for an infinite series.

Instead you'd use something like $\displaystyle N(t) = N_0(1+k)^{t}$

where
• $\displaystyle N(t)$ amount of N at time t
• $\displaystyle N_0$ amount at time 0
• $\displaystyle k$ growth rate
• $\displaystyle t$ time

Because there are 4 lots of 3 months in one year interest will be compounded 4 times so t=4.

$\displaystyle N(t) = 500(1+0.05)^4 = 607.75$

Edit:

$\displaystyle A = P\left(1 + \frac{r}{n}\right)^{nt}$

Where,
*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

The above is a more general formula for what I put.