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Math Help - Problem regarding Taylor's Theorem

  1. #1
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    Problem regarding Taylor's Theorem

    g(x) = sin(x) + cos(.5x) on (-4, 4)

    Okay, so for the above expression, I have to:
    a) Calculate an expression for the 5th derivative g^(5)(x).
    b) Find a bound for the absolute value of g^(5)(x) on the interval indicated.
    c) Use Taylor's Theorem to find a cap for the error; that is, a cap for the difference between the function and the Maclaurin polynomial of degree four on the interval.

    I did part a and this is what I got:

    g'(x) = cos(x)-.5sin(.5x)
    g''(x) = -.25cos(.5x) - sin(x)
    g'''(x) = .125sin(.5x) - cos(x)
    g^(4)(x) = .0625cos(.5x) + sin(x)
    g^(5)(x) = cos(x) - .03125sin(.5x)

    The last one is the 5th derivative.

    I attempted part b and basically what I did was substitute 4 for x in the expression for the 5th derivative of g(x). From this I got approximately .68, which I rounde up to .7. Then for a final answer, I got
    the absolute value of cos(x) - .03125sin(.5x) <= .7

    I'm not sure if the above is right, but that's what I got.

    My real problem lies in doing part c. I guess I'm confused as to the method because my book isn't very clear.
    Last edited by clockingly; March 20th 2007 at 03:33 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by clockingly View Post
    g(x) = sin(x) cos(.5x) on (-4, 4)

    Okay, so for the above expression, I have to:
    a) Calculate an expression for the 5th derivative g^(5)(x).
    b) Find a bound for the absolute value of g^(5)(x) on the interval indicated.
    c) Use Taylor's Theorem to find a cap for the error; that is, a cap for the difference between the function and the Maclaurin polynomial of degree four on the interval.

    I did part a and this is what I got:

    g'(x) = cos(x)-.5sin(.5x)
    g'(x) = -(1/2) sin(x) sin(x/2) + cos(x) cos(x/2)

    .......=3 cos(3x/2)/4 + cos(x/2)/4

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    g'(x) = -(1/2) sin(x) sin(x/2) + cos(x) cos(x/2)

    .......=3 cos(3x/2)/4 + cos(x/2)/4

    RonL

    Sorry, I forgot to add the plus sign. The expression was actually
    g(x) = sin(x) + cos(.5x) on (-4, 4)
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by clockingly View Post
    g(x) = sin(x) + cos(.5x) on (-4, 4)

    Okay, so for the above expression, I have to:
    a) Calculate an expression for the 5th derivative g^(5)(x).
    b) Find a bound for the absolute value of g^(5)(x) on the interval indicated.
    c) Use Taylor's Theorem to find a cap for the error; that is, a cap for the difference between the function and the Maclaurin polynomial of degree four on the interval.

    I did part a and this is what I got:

    g'(x) = cos(x)-.5sin(.5x)
    g''(x) = -.25cos(.5x) - sin(x)
    g'''(x) = .125sin(.5x) - cos(x)
    g^(4)(x) = .0625cos(.5x) + sin(x)
    g^(5)(x) = cos(x) - .03125sin(.5x)

    The last one is the 5th derivative.

    I attempted part b and basically what I did was substitute 4 for x in the expression for the 5th derivative of g(x). From this I got approximately .68, which I rounde up to .7. Then for a final answer, I got
    the absolute value of cos(x) - .03125sin(.5x) <= .7

    I'm not sure if the above is right, but that's what I got.

    My real problem lies in doing part c. I guess I'm confused as to the method because my book isn't very clear.
    Since |0.03125 sin(x/2)| <= 0.03125, and |cos(x)|<= 1 on (-4,4) a rough
    bound is |g^(5)(x)|<=1.03125, and the least upper bound on |g^(5)(x)|
    must be greater than or equal to 1-0.03125 = 0.96875, so 1.03125 is
    quite a good estimate of an upper bound for |g^(5)(x)|.

    RonL
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    Since |0.03125 sin(x/2)| <= 0.03125, and |cos(x)|<= 1 on (-4,4) a rough
    bound is |g^(5)(x)|<=1.03125, and the least upper bound on |g^(5)(x)|
    must be greater than or equal to 1-0.03125 = 0.96875, so 1.03125 is
    quite a good estimate of an upper bound for |g^(5)(x)|.

    RonL

    Thanks!

    I don't get why we're using just 0.03125 sin(x/2), though, instead of the entire g^5(x), though, which is cos(x)-1/32 sin (1/2 x).

    Wouldn't we just substitute 4 for x in this equation and set the equation less than or equal to -.68?

    Sorry, I'm still confused about the procedure to finding the bound.
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