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Math Help - Lagrange Error Bound Question/Problem

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    Lagrange Error Bound Question/Problem

    My Calculus teacher gave me the following problem:

    The function f has derivatives of all orders for all real numbers, and f(4) (x) = esin(x). If the third degree Taylor polynomial for f about x=0 is used to approximate f on the interval [0,1], what is the Lagrange error bound for the maximum error on the interval [0,1]?

    My peers and I had a disagreement on the correct answer and I'm curious which one of us is right.
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    Re: Lagrange Error Bound Question/Problem

    Quote Originally Posted by Xeritas View Post
    My Calculus teacher gave me the following problem:

    The function f has derivatives of all orders for all real numbers, and f(4) (x) = esin(x). If the third degree Taylor polynomial for f about x=0 is used to approximate f on the interval [0,1], what is the Lagrange error bound for the maximum error on the interval [0,1]?

    My peers and I had a disagreement on the correct answer and I'm curious which one of us is right.
    so ... what say the both of you w/ regard to the maximum error?
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    Re: Lagrange Error Bound Question/Problem

    I'm saying the Lagrange error bound should be .113 while the other individual says it should be .097.<--period

    We're using the following definition:

    |Rn(x)|<= (M/((n+1)!))|x-a|n+1
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    Re: Lagrange Error Bound Question/Problem

    for any x \in [0,1] ...

    |f(x) - P_3(x)| \le \frac{e^{\sin(1)}}{4!} \cdot x^4

    note that the error depends on the x-value in [0,1] where the function is being approximated.
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    Re: Lagrange Error Bound Question/Problem

    Quote Originally Posted by skeeter View Post
    for any x \in [0,1] ...

    |f(x) - P_3(x)| \le \frac{e^{\sin(1)}}{4!} \cdot x^4

    note that the error depends on the x-value in [0,1] where the function is being approximated.
    Yes, but the maximum error should occur furthest from the origin, which is at x=1, correct?
    And M is referring to the maximum value of the function, which would then make it:
    \frac{e^{\1}}{4!} \cdot 1^4
    Correct?
    Or is the x value within M restricted to [0,1]?
    Last edited by Xeritas; March 18th 2012 at 04:10 PM.
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    Re: Lagrange Error Bound Question/Problem

    Quote Originally Posted by Xeritas View Post
    Yes, but the maximum error should occur furthest from the origin, which is at x=1, correct?
    And M is referring to the maximum value of the function, which would then make it:
    \frac{e^{\1}}{4!} \cdot 1^4
    Correct?
    Or is the x value within M restricted to [0,1]?
    M is the maximum value of the (n+1)st derivative on the interval [0,1], M = e^{\sin(1)} in this case ... x is the specific value in [0,1] where the function is being approximated.
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    Re: Lagrange Error Bound Question/Problem

    Ok, I guess I heard wrong from our teacher, thanks again.
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