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  1. Path: sparky!uunet!wupost!waikato.ac.nz!comp.vuw.ac.nz!canterbury.ac.nz!math!wft
  2. Newsgroups: sci.math
  3. Subject: What's going on here ???
  4. Message-ID: <1992Sep11.214701.783@csc.canterbury.ac.nz>
  5. From: wft@math.canterbury.ac.nz (Bill Taylor)
  6. Date: 11 Sep 92 21:46:59 +1200
  7. Distribution: world
  8. Organization: Department of Mathematics, University of Canterbury
  9. Nntp-Posting-Host: math.canterbury.ac.nz
  10. Lines: 62
  11.  
  12. The following rather silly problem cropped up in our department just now.
  13.  
  14. Evaluate    S = 0! - 1! + 2! - 3! + 4! - ....
  15.  
  16. (yes, I know it's divergent, but blame Euler, not me.)
  17.  
  18. ------------------------
  19. Crazy solution (1)
  20.  
  21.     oo           oo     oo            oo     oo          oo
  22.    \~~   n      \~~   n /  n -x       /  -x \~~  n       /   -x 1
  23. S = > (-1) n!  = > (-1) | x e  dx  =  | e    >(-x) dx  = |  e  --- dx
  24.    /__          /__    /             /      /__         /      1+x
  25.    n=0          n=0    0             0      n=0         0
  26.  
  27.   = .5963..
  28.  
  29. -----------------------
  30. Crazy solution (2)
  31.  
  32.                                       2      3      4
  33. S = f(1)   where  f(x) = 0! - x 1! + x 2! - x 3! + x 4! - ...
  34.  
  35.                               2      3
  36. So   (x f(x))' = 1! - x 2! + x 3! - x 4! + ...
  37.  
  38.                = (1 - f(x))/x
  39.         2                                                     1        1
  40. Thus   x f' + (1+x)f = 1         this has integrating factor  - exp(- ---)
  41.                                                               x        x
  42.  
  43. Thus  [ x exp(-1/x) f(x) ]' = exp(-1/x)/x
  44.  
  45.                                x
  46.                               /
  47. thus      x exp(-x) f(x)    = | exp(-1/t)/t dt      now put x=1
  48.                               /
  49.                              0
  50.                                1
  51.                               /
  52. thus              f(1)/e  =   | exp(-1/t)/t dt      and put u=1/t
  53.                               /
  54.                              0
  55.                                oo
  56.                               /
  57.                           = e | exp(-u)/u du    =  .5963..
  58.                               /
  59.                              1
  60.  
  61. being essentially the same integral as in crazy solution (1).
  62. -----------------------
  63.  
  64. So what's going on here ????
  65.  
  66. Two crazy solutions to a crazy problem giving the same answer.   Why ?
  67.  
  68. --------------------------------------------------------------------------
  69.             Bill Taylor              wft@math.canterbury.ac.nz
  70. --------------------------------------------------------------------------
  71.  MATH: necessary consequences of arbitrary axioms about meaningless things.
  72. --------------------------------------------------------------------------
  73.  
  74.