On page 95 of Zima-Brown you will find a formula that gives a starting value for finding the solution i to the equation sn|i = k. The formula is
Where does this formula come from?
The equation we want to solve (for i) is k = ((1+i)n-1)/i. In situations like this, we want to approximate the complicated term (1+i)n by a simpler expression in i and n so that solving for i becomes possible.
The obvious approach is to use the binomial theorem and write (1+i)n = 1+ni+(n(n-1)/2)i2+¼. If we plug this into the formula for sn|i we get the equation k = n+(n(n-1)/2)i+¼ and if we ignore the '¼' terms we get a simple equation whose solution is i » 2(k-n)/(n(n-1)). If you test this formula, you will find that it is not as good as the one given in the book! The reason is that although the dropped terms involve powers of i like i2 and i3 and so on and these powers are indeed very small, the coefficients of these dropped terms are large: n(n-1)(n-2)/6, n(n-1)(n-2)(n-3)/24 etc. Hence a significant error is made when we drop these terms. Thus the obvous approach fails!
The difficulty is that terms of the size ni may not be small in the cases of importance to us. For example, if n = 10 years and i = 0.10 per year, then ni = 1. Or if n = 100 months and i = 0.01 per month then ni = 1 again.
One thing to try, when the obvious approach fails, is a difference of two binomial expansions where terms will cancel (or partially cancel) each other. With that in mind let us write
The term in square brackets is a difference of binomial expansions. When we expand both parts, some cancellation occurs: the 1's cancel; the next terms are ni/2 and (-n)i/2 and these subtract to give ni; the next terms are (n/2)((n/2)-1)i2/6 and (-n/2)((-n/2)-1)i2/6 and these subtract to give -ni2/6. This last term is safe to drop! So are the remaining terms because of partial cancellations and because the factorial denominators of the binomial expansion are starting to become large. (The next denominator is 4! = 24, then 5! = 120 and so on.)
Hence we have an approximate equality: sn|i » (1+i)n/2[ni]/i = (1+i)n/2n. Thus we can say k » n(1+i)n/2 or k/n » (1+i)n/2. Squaring this gives (k/n)2 » (1+i)n and putting this into our equation to be solved gives an equation k » ((k/n)2-1)/i so i » ((k/n)2-1)/k as required.