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 Initial Term Identification (Posted on 2015-07-21)
A sequence {X0, X1, X2, ...., Xn} is given by:
Xn(1 - Xn-1) = Xn-1

Find X0, given that: X2015 = 2015

 No Solution Yet Submitted by K Sengupta No Rating

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 electronically computed | Comment 1 of 3
x(n) - x(n)*x(n-1) = x(n-1)
x(n-1)*(1 + x(n)) = x(n)
x(n-1) = x(n)/(1+x(n))

So

4    kill "initterm.txt"
5    open "initterm.txt" for output as #2
10   X=2015:Lvl=2015
20   for Lvl=2014 to 0 step -1
30    X=X//(1+X)
35   print #2,Lvl,X
40   next
45   close #2
50   print X

In the first few terms the numerator remains 2015, but the denominator, initially 1, goes up by 2015 each successively smaller subscript

```  n        x(n)
2014   2015//2016  2013   2015//4031  2012   2015//6046  2011   2015//8061  2010   2015//10076  2009   2015//12091  2008   2015//14106  2007   2015//16121  2006   2015//18136  2005   2015//20151    ...  and the last few are:  ...  9   2015//4042091  8   2015//4044106  7   2015//4046121  6   2015//4048136  5   2015//4050151  4   2015//4052166  3   2015//4054181  2   2015//4056196  1   2015//4058211  0   2015//4060226 ```

So x(0) = 2015/4060226

 Posted by Charlie on 2015-07-21 15:08:06

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