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Thread: hay guyz

  1. #1

    hay guyz

    Use Taylor's Theorem to obtain an upper bound for the error of the approximation. Then calculate the exact value of the error.

    arctan(0.4) ≈ 0.4 - (0.4)³/3

    So it's screwing with my head.

  2. #2
    Old school, like an old fool. Flying Mullet's Avatar
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    I'm assuming that you're using a Taylor Polynomial of degree 3:
    x - x³/3

    To find the upper bound, or remainder, use the following formula:
    ( max | f(n+1)(z) | * (x – x0) n+1 ) / (n + 1)!

    For you, f(x) is arctan(x), n is 3 (your Taylor Polynomial is up to x³) and x0 is 0, making your equation look like this:
    ( max | f(4)(z) | * (x)^4 ) / (4)!

    You'll need to find the fourth derivative of arctan and then find the maximum value for the fourth derivative of arctan over the interval [0, 0.4].

    The fourth derivative of arctan is x(6x^2+9)/(1-x^2)^(7/2). The max value on [0, 0.4] is 7.334.

    So your final formula is:
    ((7.334)x^4) / (4)!

    Substitute 0.4 in for x and you get:
    ((7.334)(0.4)^4) / (4)! = 0.007823

    The forumla for the exact remainder is:
    | f(x) - Pn(x)| or
    | arctan(x) - x - x^3/3 |

    So subtiitute 0.4 in for x again and you get:
    | arctan(0.4) - (0.4) 0 (0.4)^3/3 | = 0.001833

    Thus 0.007823 is the upper bound of the error and 0.001833 is the exact remainder.




    You wouldn't happen to be doing problem #48 in section 9.7 of <a href="http://www.amazon.com/Calculus-Single-Variable-Transcendental-Functions/dp/0618606254/ref=sr_11_1?ie=UTF8&qid=1195139466&sr=11-1">this book</a>, would you?
    Figaro Castle

  3. #3
    Mr. Encyclopedia Kirobaito's Avatar
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    It amazes me that a year and a half ago I would have been able to do this problem. It's not hard at all. I remember knowing how to do Taylor polynomials. But it's all gone. I wouldn't have the first clue what to do.

  4. #4
    Toy with and destroy xXsarahXx's Avatar
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    what in the name of god is that?!?!

  5. #5

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    Quote Originally Posted by Flying Mullet View Post
    I'm assuming that you're using a Taylor Polynomial of degree 3:
    x - x³/3

    To find the upper bound, or remainder, use the following formula:
    ( max | f(n+1)(z) | * (x – x0) n+1 ) / (n + 1)!

    For you, f(x) is arctan(x), n is 3 (your Taylor Polynomial is up to x³) and x0 is 0, making your equation look like this:
    ( max | f(4)(z) | * (x)^4 ) / (4)!

    You'll need to find the fourth derivative of arctan and then find the maximum value for the fourth derivative of arctan over the interval [0, 0.4].

    The fourth derivative of arctan is x(6x^2+9)/(1-x^2)^(7/2). The max value on [0, 0.4] is 7.334.

    So your final formula is:
    ((7.334)x^4) / (4)!

    Substitute 0.4 in for x and you get:
    ((7.334)(0.4)^4) / (4)! = 0.007823

    The forumla for the exact remainder is:
    | f(x) - Pn(x)| or
    | arctan(x) - x - x^3/3 |

    So subtiitute 0.4 in for x again and you get:
    | arctan(0.4) - (0.4) 0 (0.4)^3/3 | = 0.001833

    Thus 0.007823 is the upper bound of the error and 0.001833 is the exact remainder.




    You wouldn't happen to be doing problem #48 in section 9.7 of <a href="http://www.amazon.com/Calculus-Single-Variable-Transcendental-Functions/dp/0618606254/ref=sr_11_1?ie=UTF8&qid=1195139466&sr=11-1">this book</a>, would you?
    Thanks a bunch

    And yeah, it's #47 in 9.7, but it's not Early Transcendental Functions, it's just Calculus of A Single Variable, Eighth Edition. But it's the same number and section, so I don't know how different they are.

  6. #6
    Old school, like an old fool. Flying Mullet's Avatar
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    If you run into any more problems, check out this site:
    Calc Chat Free Solutions

    It's the online companion to the textbook and they worked out the solutions to the odd number problems there.

    Hopefully if you're scratching your head on a problem it will help you.
    Figaro Castle

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