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Alternating series, error estimation & approximation

  • Thread starter johnnyies
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  • #1
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Homework Statement


[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.

Homework Equations


No idea.


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.
 

Answers and Replies

  • #2
vela
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What does the alternating series theorem tell you?
 
  • #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
 
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  • #4
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so I just plug in numbers?
 
  • #5
vela
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Yeah.
 
  • #6
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. . . that doesn't even seem viable to me. . . it's essentially guessing until you get the right error?
 
  • #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.
 
  • #8
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the magnitude of the error of n terms is less than the next n + 1 th term?
 
  • #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.
 
  • #10
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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?
 
  • #11
vela
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Yes, but n=6 also works, right?
 
  • #12
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ya, much thanks!
 

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