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Math Help - Taylor Series

  1. #1
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    Taylor Series

    Find the Taylor Polynomial P_n(x) for the function f with the given values of c and n. Then give a bound on the error that is incurred if P_n(x) is used to approximate f(x) on the given interval. f(x)=e^(3x) c=1, n=4, [1, 1.1]

    I found P_4(x):

    P_4(x)=(e^3) + 3(e^3)(x-1) + (9/2)(e^3)(x-1)^2 + (9/2)(e^3)(x-1)^3 + (27/8)(e^3)(x-1)^4

    but the problem is also asking for the absolute value of R_4(x) less than or equal to...

    How do I find R_4(x)? The answer they gave is .000549
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  2. #2
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    Re: Taylor Series

    Lagrange form of the remainder ...

    |R_n(x)| \le \, max|f^{(n+1)}(z)| \cdot \frac{|x-c|^{n+1}}{(n+1)!}

    R_4(x) \le \frac{81e^{3.3}(1.1 - 1)^5}{40} = 0.000549
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    Re: Taylor Series

    Quote Originally Posted by skeeter View Post
    Lagrange form of the remainder ...

    |R_n(x)| \le \, max|f^{(n+1)}(z)| \cdot \frac{|x-c|^{n+1}}{(n+1)!}

    R_4(x) \le \frac{81e^{3.3}(1.1 - 1)^5}{40} = 0.000549
    Where did the 3.3 come from? And I thought that form was saying to use the 5th derivative? Or are you really just using the 4th derivative?
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  4. #4
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    Re: Taylor Series

    R_4(x) is the remainder of a 4th degree Taylor polynomial.

    |R_4(x)| \le \, max|f^5(z)| \cdot \frac{|x-1|^5}{5!}


    243e^{3.3} is the maximum possible value of f^5(z) , where z \in [1,1.1]
    Thanks from Preston019
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    Re: Taylor Series

    Quote Originally Posted by skeeter View Post
    R_4(x) is the remainder of a 4th degree Taylor polynomial.

    |R_4(x)| \le \, max|f^5(z)| \cdot \frac{|x-1|^5}{5!}


    243e^{3.3} is the maximum possible value of f^5(z) , where z \in [1,1.1]
    Oh! Okay, I see what i was doing wrong: I was plugging in 1 to e^3x instead of 1.1

    Thank You!
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