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Math Help - taylor series

  1. #1
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    taylor series

    how would you do this?
    Find the Taylor series for ln(x) about 2.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by davecs77 View Post
    how would you do this?
    Find the Taylor series for ln(x) about 2.
    see here

    if you need further explanation, just say so

    topsquark showed you the main process of deriving something like this, but again, memorization could be used here, the page i gave you has the general formulas
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by davecs77 View Post
    how would you do this?
    Find the Taylor series for ln(x) about 2.
    Just for the record, let f(x) = ln(x)

    f(x) = ln(x) ==> f(2) = ln(2)

    f^{\prime}(x) = \frac{1}{x} ==> f^{\prime}(2) = \frac{1}{2}

    f^{\prime \prime}(x) = -\frac{1}{x^2} ==> f^{\prime \prime}(2) = -\frac{1}{4}

    etc.

    So
    f(x) \approx ln(2) + \frac{1}{1!}\frac{1}{2}(x - 2) - \frac{1}{2!}\frac{1}{4}(x - 2)^2 + ~ ...

    f(x) \approx ln(2) + \frac{1}{2}(x - 2) - \frac{1}{8}(x - 2)^2 + ~ ...

    -Dan
    Last edited by topsquark; July 23rd 2007 at 04:25 PM. Reason: Fixed an error in the form of the expansion
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  4. #4
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    Quote Originally Posted by topsquark View Post
    Just for the record, let f(x) = ln(x)

    f(x) = ln(x) ==> f(2) = ln(2)

    f^{\prime}(x) = \frac{1}{x} ==> f^{\prime}(2) = \frac{1}{2}

    f^{\prime \prime}(x) = -\frac{1}{x^2} ==> f^{\prime \prime}(2) = -\frac{1}{4}

    etc.

    So
    f(x) \approx ln(2) + \frac{1}{1!}\frac{1}{2}x - \frac{1}{2!}\frac{1}{4}x^2 + ~ ...

    f(x) \approx ln(2) + \frac{1}{2}x - \frac{1}{8}x^2 + ~ ...

    -Dan
    How did you get 1/1! times (1/2)(x) - 1/2! times (1/40(x^2)
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  5. #5
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    Quote Originally Posted by davecs77 View Post
    How did you get 1/1! times (1/2)(x) - 1/2! times (1/40(x^2)

    The answer in my book is ln2 + summation of n = 1 going to infinity of
    (-1)^(n - 1) times ((x - 2)^n)/(n2^n)

    I get the first part..ln2 but dont understand the rest. Thank you.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by davecs77 View Post
    How did you get 1/1! times (1/2)(x) - 1/2! times (1/40(x^2)
    Crap! I goofed!

    The general form for the Taylor series of f(x) about x = a is:
    f(x) \approx f(a) + \frac{1}{1!}f^{\prime}(a)(x - a) + ~ ... ~ + \frac{1}{n!}f^{(n)}(a)(x - a)^n + ~ ...

    So for a = 2 the first coefficient is
    \frac{1}{1!}f^{\prime}(2) = 1 \cdot \frac{1}{2} = \frac{1}{2}

    The second coefficient is
    \frac{1}{2!}f^{\prime \prime}(2) = \frac{1}{2} \cdot -\frac{1}{4} = -\frac{1}{8}

    So the series should be:
    f(x) \approx ln(2) + \frac{1}{2}(x - 2) - \frac{1}{8}(x - 2)^2 + ~ ...

    You can work the general formula for the nth term. It's easy enough.

    -Dan
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    Crap! I goofed!
    can a Mormon say that? you know, the C-word, i only quoted "goofed" because it was funny as h
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  8. #8
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    Quote Originally Posted by topsquark View Post
    Crap! I goofed!

    The general form for the Taylor series of f(x) about x = a is:
    f(x) \approx f(a) + \frac{1}{1!}f^{\prime}(a)(x - a) + ~ ... ~ + \frac{1}{n!}f^{(n)}(a)(x - a)^n + ~ ...

    So for a = 2 the first coefficient is
    \frac{1}{1!}f^{\prime}(2) = 1 \cdot \frac{1}{2} = \frac{1}{2}

    The second coefficient is
    \frac{1}{2!}f^{\prime \prime}(2) = \frac{1}{2} \cdot -\frac{1}{4} = -\frac{1}{8}

    So the series should be:
    f(x) \approx ln(2) + \frac{1}{2}(x - 2) - \frac{1}{8}(x - 2)^2 + ~ ...

    You can work the general formula for the nth term. It's easy enough.

    -Dan
    You are only doing 2 coefficients since a = 2?
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by davecs77 View Post
    You are only doing 2 coefficients since a = 2?
    No, I only did two terms since I'm lazy. The series is infinite with each term being smaller than the last.

    -Dan
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  10. #10
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    Quote Originally Posted by davecs77 View Post
    how would you do this?
    Find the Taylor series for ln(x) about 2.
    An indirect way to find the taylor series is to use the following known Maclaurin series
    Attached Thumbnails Attached Thumbnails taylor series-july10.gif  
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  11. #11
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    Quote Originally Posted by curvature View Post
    An indirect way to find the taylor series is to use the following known Maclaurin series
    Then
    Attached Thumbnails Attached Thumbnails taylor series-july11.gif  
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