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Use taylor's THeorem to determine the accuracy

  1. Apr 29, 2009 #1
    Use taylor's THeorem to determine the accuracy of the approximation: cos(.3) ~=1 - (.3)^2 / 2! + (.3)^4 / 4! when i use taylors theorem, i use (.3)^4 / 4! which gets me 2.03e-10 but the asnwer is R<=2.03e-5
     
  2. jcsd
  3. Apr 29, 2009 #2

    Dick

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    Re: Series

    (.3)^4/4! is not equal to 2.03e-10. What form of the remainder term are you using?
     
  4. Apr 29, 2009 #3
    Re: Series

    Okay, it should be (.3)^5 / 5!
     
    Last edited: Apr 29, 2009
  5. Apr 29, 2009 #4
    Re: Series

    But i still dont get it, why did i get hte right answer and ignore the f^(n+1)(c) of the formula? (Lagrange error bound) The derivative should have been cos , so i must have picked cos 0. Why?
     
  6. Apr 30, 2009 #5

    HallsofIvy

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    Re: Series

    The error does NOT involve f^(n+1)(c). It involves f^(n+1)(x) for some number between 0 and c (here .3). Since you don't know that number, you use the largest possible value of f^(n+1)(x) between 0 and c to get an upper bound on the error. Since cosine is decreasing, its largest value is cos(0)= 1.
     
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