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Alternating series, error estimation & approximation

  1. Apr 2, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex]\Sigma[/tex](-1)[tex]^{n+1}[/tex][tex]\frac{1}{n!}[/tex]

    How many terms will suffice to get an approximation within 0.0005 of the actual sum? Find that approximation.
    2. Relevant equations
    No idea.


    3. The attempt at a solution
    What I tried doing is setting my absolute value of the series less than 0.005, but I have no idea how to get rid of that factorial.
     
  2. jcsd
  3. Apr 2, 2010 #2

    vela

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    What does the alternating series theorem tell you?
     
  4. Apr 2, 2010 #3

    lanedance

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    as the magnitude of the terms are monotonically decreasing, and alternating, you could also look at the magnitude of a single term
     
    Last edited: Apr 2, 2010
  5. Apr 2, 2010 #4
    so I just plug in numbers?
     
  6. Apr 2, 2010 #5

    vela

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    Yeah.
     
  7. Apr 2, 2010 #6
    . . . that doesn't even seem viable to me. . . it's essentially guessing until you get the right error?
     
  8. Apr 2, 2010 #7

    vela

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    Well, you shouldn't be making a wild guess. What exactly are you trying to do? You never answered my question about what you know about alternating series, in particular, about the error.
     
  9. Apr 3, 2010 #8
    the magnitude of the error of n terms is less than the next n + 1 th term?
     
  10. Apr 3, 2010 #9

    vela

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    Right, so you're trying to solve

    [tex]\frac{1}{(n+1)!} < 0.0005[/tex]

    Find how big (n+1)! has to be and then what n would satisfy that.
     
  11. Apr 3, 2010 #10
    invert both sides with inequality switched

    (n+1)! > 2000

    so (6+1)! = 5040 > 2000

    so all n > 6 will make an error less than 0.0005?
     
  12. Apr 3, 2010 #11

    vela

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    Yes, but n=6 also works, right?
     
  13. Apr 3, 2010 #12
    ya, much thanks!
     
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