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Math Help - Taylor Series and Approximation

  1. #1
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    Taylor Series and Approximation

    Hello Guys,

    I need help solving these two problem, in the first one i dont really know what im supposed to do with the x = 9... Please i need help...

    Problem 1

    Find the Taylor series for f(x) = √x using reference point
    x = 9. For what values of x does it converge to √x?


    Problem 2

    Estimate ln(1/2) to within 1/1000.
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  2. #2
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    Quote Originally Posted by tapiaghector View Post
    Hello Guys,

    I need help solving these two problem, in the first one i dont really know what im supposed to do with the x = 9... Please i need help...

    Problem 1

    Find the Taylor series for f(x) = √x using reference point
    x = 9. For what values of x does it converge to √x?


    Problem 2

    Estimate ln(1/2) to within 1/1000.
    f'(x)=\frac{1}{2}x^{-1/2}
    f''(x)=\frac{-1}{2^2}x^{-3/2}
    f'''(x)=\frac{3}{2^3}x^{-5/2}
    f^{4}(x)=\frac{-3\cdot 5}{2^4}x^{-7/2}

    for n \ge 2
    f^{n}(x)=\frac{(-1)^{n+1}1\cdot 3 \cdots (2n-3)}{2^n}x^{\frac{-(2n-1)}{2}}

    f^{n}(9)=\frac{(-1)^{n+1}1\cdot 3 \cdots (2n-3)}{2^n}(9)^{\frac{-(2n-1)}{2}}

    f^{n}(9)=\frac{(-1)^{n+1}1\cdot 3 \cdots (2n-3)}{3^{2n-1}2^n}

    T(x)=3+\frac{1}{6}(x-3)+\sum_{n=2}^{\infty}\frac{(-1)^{n+1}1\cdot 3 \cdots (2n-3)}{2^n 3^{2n-1}n!}(x-9)^n

    For the 2nd start with

    f(x)=ln(x+1)

    f'(x)=\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n

    Now we can integrate to get

    f(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{n+1}}{n+1}

    You want

    f\left( \frac{1}{2}\right)=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}n+1}

    This is an alternating series so the error cannot be bigger then the next term in the series n=10 will work because 2^{10}=1024
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  3. #3
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    Hey TheEmptySet,

    Thanks for your help, know i understand more about taylor series...

    Hava a great night and thanks again!
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