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Math Help - Maclaurin series help

  1. #1
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    Maclaurin series help



    Last edited by Amanda-UK; August 22nd 2007 at 04:38 AM. Reason: missed a bit
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  2. #2
    Forum Admin topsquark's Avatar
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    A Maclauren series for a function is just the Taylor series at an argument of 0.

    So for f(x) = ln(1 + x):
    f(x) = ln(1 + x) \implies f(0) = 0

    f^{\prime}(x) = \frac{1}{1 + x} \implies f^{\prime}(0) = 1

    f^{\prime \prime}(x) = -\frac{1}{(1 + x)^2} \implies f^{\prime \prime}(0) = -1

    etc.

    So
    f(x) \approx f(0) + \frac{1}{1!}f^{\prime}(0)x + \frac{1}{2!}f^{\prime \prime}(0)x^2 + ~ ...

    f(x) \approx 0 + \frac{1}{1} \cdot 1 \cdot x + \frac{1}{2} \cdot (-1) \cdot x^2 + ~ ...

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

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    A Maclauren series for a function is just the Taylor series at an argument of 0.

    So for f(x) = ln(1 + x):
    f(x) = ln(1 + x) \implies f(0) = 0

    f^{\prime}(x) = \frac{1}{1 + x} \implies f^{\prime}(0) = 1

    f^{\prime \prime}(x) = -\frac{1}{(1 + x)^2} \implies f^{\prime \prime}(0) = -1

    etc.

    So
    f(x) \approx f(0) + \frac{1}{1!}f^{\prime}(0)x + \frac{1}{2!}f^{\prime \prime}(0)x^2 + ~ ...

    f(x) \approx 0 + \frac{1}{1} \cdot 1 \cdot x + \frac{1}{2} \cdot (-1) \cdot x^2 + ~ ...



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

    -Dan
    would you be able to go into more detail wit this please?also stuck on B
    thank u
    Amandax
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    Now the second one show that f(1) and f(2) have different signs. That means, by intermediate value theorem there is a number x_0\in (1,2) so that f(x_0)=0.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Amanda-UK View Post
    would you be able to go into more detail wit this please?
    Amandax
    More detail with what? What specifically are you having a problem with?

    -Dan
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