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  1. Newsgroups: sci.math.num-analysis
  2. Path: sparky!uunet!cs.utexas.edu!torn!skule.ecf!rairan
  3. From: rairan@ecf.toronto.edu (RAI Ranjan)
  4. Subject: Re: *moan*beg*: HELP : sob,snif "forlorn looks", much suffering....
  5. Message-ID: <BtGyLy.9EB@ecf.toronto.edu>
  6. Keywords: COMMUNICATION BREAKDOWN ...
  7. Organization: University of Toronto, Engineering Computing Facility
  8. References: <1992Aug23.113011.816@ecst.csuchico.edu>
  9. Distribution: usa
  10. Date: Mon, 24 Aug 1992 03:49:10 GMT
  11. Lines: 30
  12.  
  13. In article <1992Aug23.113011.816@ecst.csuchico.edu> madams@ecst.csuchico.edu (Michael E. Adams) writes:
  14. >Given:  p, n, b, & f = 0;  How do you solve for 'i'?
  15. >
  16. >                            -n                       -n
  17. >     100 p  1 - (1 + i/100)^   +  b = -f (1 + i/100)^
  18. >            ---------------
  19. >                   i
  20.  
  21.  
  22.    To all the guys who responded to this original posting:  Mike DIDN'T mean
  23.  p = n = b = f = 0.  Only f = 0.  You would think the guy from Stanford would
  24.  have figured that out :)  
  25.  
  26.    Anyway, there is no general `formula' for `i' in terms of the other 
  27.  quantities, but you can make some considerable simplifications to the given
  28.  equation: 
  29.                                         -n                   
  30.                         1 - (1 + i/100)^               
  31.                    100p ---------------    + b = 0  
  32.                               i
  33.    let y = 1 + i/100 and after some manipulation you get the much nicer eqn.
  34.            (b/p)*y^(n+1) + (1-b/p)*y^(n) = 1, which is a polynomial.
  35.  
  36.    Now use whatever root-finding code you may have (remember i !=0 ).
  37.  
  38. -- 
  39. ---
  40. Ranjan Rai                             <rairan@ecf.toronto.edu> 
  41. Engineering Science 9T3                <rairan@eecg.toronto.edu>
  42. University of Toronto                      
  43.