🚀 go-pugleaf

In article <1992Aug23.113011.816@ecst.csuchico.edu> madams@ecst.csuchico.edu (Michael E. Adams) writes:
>Given:  p, n, b, & f = 0;  How do you solve for 'i'?
>
>                            -n                       -n
>     100 p  1 - (1 + i/100)^   +  b = -f (1 + i/100)^
>            ---------------
>                   i


   To all the guys who responded to this original posting:  Mike DIDN'T mean
 p = n = b = f = 0.  Only f = 0.  You would think the guy from Stanford would
 have figured that out :)  

   Anyway, there is no general `formula' for `i' in terms of the other 
 quantities, but you can make some considerable simplifications to the given
 equation: 
                                        -n                   
                        1 - (1 + i/100)^               
                   100p ---------------    + b = 0  
                              i
   let y = 1 + i/100 and after some manipulation you get the much nicer eqn.
           (b/p)*y^(n+1) + (1-b/p)*y^(n) = 1, which is a polynomial.

   Now use whatever root-finding code you may have (remember i !=0 ).

-- 
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Ranjan Rai                             <rairan@ecf.toronto.edu> 
Engineering Science 9T3                <rairan@eecg.toronto.edu>
University of Toronto