sorry, it's been a while. Mitch asked for our equation, so here's our thought process on how we got to it from 30% yield and 70% recovered SM

iteration 1 yield: 1 x 0.3

iteration 2 yield: (1 x 0.7) x 0.3

iteration 3 yield: ((1 x 0.7)0.7) x 0.3

iteration 4 yield: (((1 x 0.7)0.7)0.7) x 0.3

...

thus, total yield for all iterations:

total yield = 0.30[ (1) + (1 x 0.7) + (1 x 0.7

^{2}) + (1 x 0.7

^{3}) ]

total yield = 0.30[ (1 x 0.7

^{0}) + (1 x 0.7

^{1}) + (1 x 0.7

^{2}) + (1 x 0.7

^{3}) ]

total yield = 0.30(0.7

^{0} + 0.7

^{1} + 0.7

^{2} + 0.7

^{3})

or

Σ

_{x=0→n} y(1-y)

^{x} (where y is the yield for each iteration and n is the number of iterations)

or more generally for any recycling experiment where the mass balance is not 100%

Σ

_{x=0→n} y(z)

^{x} (where y is the yield for each iteration and z is the percent recovered starting material and n is the number of iterations)

There might be a way to convert the summation into a function, but I don't remember my calculus that well.