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LaGrange Remainder Infinite Series

  1. Mar 28, 2009 #1
    1. The problem statement, all variables and given/known data

    Let f be a function whose seventh derivative is f7(x) = 10,000cos x. If x = 1 is in the interval of convergence of the power series for this function, then the Taylor polynomial of degree six centered at x = 0 will approximate f(1) with an error of not more than

    a.) 2.45 x 10-5

    b.) 1.98 x 10-4

    c.) 3.21 x 10-2

    d.) 0.248

    e.) 1.984

    2. Relevant equations

    The Lagrange Remainder Formula, it states that the biggest error is only as large as the next sum in the series.

    The formula is:

    Rn [tex]\leq[/tex][tex]\frac{f^{7}(z)(x-c)^7}{(n+1)!}[/tex]

    3. The attempt at a solution

    The maximum error is the next term in the sequence. Looking at the lagrange formula, we're looking for the maximum error, or the f7(z) term. Since we are given f7(x) = 10,000cos x and given that x = 1 is in the interval of convergence... I assumed that f7(1) = 10,000cos(1) is the maximum error.

    Plugging it back in the LaGrange Formula I get the following:

    R6 [tex]\leq[/tex] [tex]\frac{f^{7}(z)x^7}{7!}[/tex] = [tex]\frac{10000cos 1}{7!}[/tex]

    I get a number that is not in the multiple choice answers. Any tips/ideas? Thanks
     
  2. jcsd
  3. Mar 28, 2009 #2

    HallsofIvy

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    Science Advisor

    Why would you assume that? Cosine is a decreasing function between 0 and 1< [itex]\pi/2[/itex]. It takes its largest value at the lowest value of x in the interval, not the highest.

     
  4. Mar 28, 2009 #3
    I see my error. What threw me off is at the beginning of the problem how "1" is in the interval of convergence. I immediately thought that f7 (1) was the maximum value.

    We're just looking at the maximum value of f(x) = 10,000cos(x) correct? and cosine is between -1 and 1... so the maximum value is cos(0).

    The lagrange remainder would be 10,000 / 7!.

    Is this correct?
     
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