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Math Help - Standard Taylor series

  1. #1
    Junior Member froodles01's Avatar
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    Post Standard Taylor series

    Question is: Using a standard Taylor series, write down the Taylor series about 0 for the function

    f(x) = ln(1 x),
    giving all terms up to and including that include x^4
    Now, Standard Taylor series of ln(1+x), I understand, is;

    ln(1+x) = x - (1/2 x^2) +(1/3 x^3) - (1/4 x^4) + . . . . .
    so substitute -x for +x I get

    ln(1-x) = -x -(-1/2 x^2) + (-1/3 x^3) - (-1/4 x^4) + . . . .
    Ln(1-x) = -x + (x^2)/2 - (x^3)/3 + (x^4)/4 - . . . . . .

    But the answer I have been given as a specimen makes all of the signs -ve
    Could someone clarify this, please.

    I always assume I'm wrong, but rarely see why.
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  2. #2
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    Re: Standard Taylor series

    Can you simplify \frac{1}{2}(-x)^2?
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  3. #3
    Junior Member froodles01's Avatar
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    Re: Standard Taylor series

    -(x^2)/2
    Yes!
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    Re: Standard Taylor series

    You need to review arithmetic with negative numbers, e.g., this page: "Why is the Product of Negative Numbers Positive?"
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    Re: Standard Taylor series

    Quote Originally Posted by froodles01 View Post
    Question is: Using a standard Taylor series, write down the Taylor series about 0 for the function

    f(x) = ln(1 x),
    giving all terms up to and including that include x^4
    Now, Standard Taylor series of ln(1+x), I understand, is;

    ln(1+x) = x - (1/2 x^2) +(1/3 x^3) - (1/4 x^4) + . . . . .
    so substitute -x for +x I get

    ln(1-x) = -x -(-1/2 x^2) + (-1/3 x^3) - (-1/4 x^4) + . . . .
    Ln(1-x) = -x + (x^2)/2 - (x^3)/3 + (x^4)/4 - . . . . . .

    But the answer I have been given as a specimen makes all of the signs -ve
    Could someone clarify this, please.

    I always assume I'm wrong, but rarely see why.
    you have made a mistake. Note that in log(1+x) all the even powers of x are - terms and all odd powers of x are + terms So when you replace x by -x even powers will stay - and odd powers will become -
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