Lets assume we have a mikvah of 40 seah kosher (rainwater) & 10 seah posul (pail water)

Now we add 1/37 (lets call this 'w') seah of posul,then we draw out the same amount from the whole mixture.(also known as ''נתן סאה ונטל סאה").Imagine you repeat this again & again.

We wish to find what is the amount of 'kosher'('k') water remaining in mikvah after repeating this process 5 times ('n' times)

When adding the posul water, we have in total 50+w ,40 k & 10+w p.Then we draw w ,how much of w is k? we find this by w*(40/(50+w)) {the fraction of k water from total }

so we finally have 40-{w*(40/(50+w))};this is the amount of k water left in pool after first process.

lets simplify the expression in { }is =(w*40)/(50+w ),or 40 * (w/(50+w)). lets call this last fraction 'y'.

we know that 40-40y= 40(1-y),so we finally see that :40- 40 * y= 40*(1-y). this is the amount of kosher water remaining after first process. so we just repeat this for every 'action'נתינה ונטילה

second action : kosher or k= 40 *(1-y)*(1-y)....

or k=40*(1-y)^a .('a' is 'action')

example: after 7 actions how much kosher water is left?

40* {1-(.027027/50.027027)}^7=40 *.999459752^7=39.84897548...

to generalize:if t is total water you start off (in this example it is 50) & you start with q(in this example 40)kosher

the formula is thus:q* {1-(w/(t+w))}^a