Half life of a thing is 'h' days. I have a box where everyday I put 'n' quantity into the box. How do I calculate the quantity remaining after 'd' days.
i.e. if half life is 7 days. Each day I add 10 gms of the item to the box, then on the 7th day 7th day, just before I add the 10 gms for the day, I have
N = 10 * (1/2)^ (7/7) + 10 * (1/2) ^(6/7) + 10* (1/2) ^(5/7) + ....
N = 5 + 5.5 + 6.1 + 6.7 + 7.4 + 8.2 + 9.1
N = 48
Can a formula be derived from this into which I can plug in h, n & d & it gives me the quantity remaining
Lets rewrite your sum in general terms $$N = 10 \cdot (1/2)^{(7/7)} + 10 \cdot (1/2) ^{(6/7)} + 10 \cdot (1/2) ^{(5/7)}+\ldots$$
First note, that the factor between each term is $(1/2)^{(1/7)}$, the first term is $\left((1/2)^{(1/7)}\right)^7$, the second term is $\left((1/2)^{(1/7)}\right)^6$ and so on.
In order to express this more easily in variables, i introduce this factor as $v$:
Let $v = 2^{-\frac{1}{h}}$.
$v$ is the fraction remaining of one unit after one day. Then $$N = \sum_{k=1}^{d} (n\cdot v^k)$$ The first term in the sum is the remaining amount of yesterday, ... the last term of the sum is the amount remaining from $d$ days ago.
Now you can use the formula for geometric series